An intuitive understanding of continuity - AP Calculus AB
Card 1 of 33
$f(x)={x^{2}$+3} when 
and
f(x)={7x-3} when xgeq 3
At 
the funciton described above is:
$f(x)={x^{2}$+3} when
f(x)={7x-3} when xgeq 3
At
Tap to reveal answer
The answer is both.
If graphed the student will see that the two graphs are continuous at 
. There is no gap in the graph or no uneven transitions. If the graph is continuous then it is differentiable so it must be both.
The answer is both.
If graphed the student will see that the two graphs are continuous at
← Didn't Know|Knew It →
Which of the following functions contains a removeable discontinuity?
Which of the following functions contains a removeable discontinuity?
Tap to reveal answer
A removeable discontinuity occurs whenever there is a hole in a graph that could be fixed (or "removed") by filling in a single point. Put another way, if there is a removeable discontinuity at 
, then the limit as 
approaches 
exists, but the value of 
does not.
For example, the function $f(x)=\frac{1+x^3$}{1+x}$ contains a removeable discontinuity at 
. Notice that we could simplify
as follows:
$f(x)=\frac{1+x^3$$}{1+x}$=\frac{(1+x)(x^2$$-x+1)}{1+x}$=x^{2}$-x+1, where xneq -1.
Thus, we could say that $lim_{xrightarrow $-1}$$\frac{1+x^$3$}{1+x}$=lim_{xrightarrow $-1}$x^2$$-x+1=(-1)^2$-(-1)+1=3.
As we can see, the limit of 
exists at 
, even though 
is undefined.
What this means is that 
will look just like the parabola with the equation $x^{2}$-x+1 EXCEPT when
, where there will be a hole in the graph. However, if we were to just define 
, then we could essentially "remove" this discontinuity. Therefore, we can say that there is a removeable discontinuty at 
.
The functions
, and
have discontinuities, but these discontinuities occur as vertical asymptotes, not holes, and thus are not considered removeable.
The functions
and $f(x)=\frac{x+1}{1+x^{2}$$} are continuous over all the real values of 
; they have no discontinuities of any kind.
The answer is
.
A removeable discontinuity occurs whenever there is a hole in a graph that could be fixed (or "removed") by filling in a single point. Put another way, if there is a removeable discontinuity at
For example, the function $f(x)=\frac{1+x^3$}{1+x}$ contains a removeable discontinuity at
$f(x)=\frac{1+x^3$$}{1+x}$=\frac{(1+x)(x^2$$-x+1)}{1+x}$=x^{2}$-x+1, where xneq -1.
Thus, we could say that $lim_{xrightarrow $-1}$$\frac{1+x^$3$}{1+x}$=lim_{xrightarrow $-1}$x^2$$-x+1=(-1)^2$-(-1)+1=3.
As we can see, the limit of
What this means is that
The functions
have discontinuities, but these discontinuities occur as vertical asymptotes, not holes, and thus are not considered removeable.
The functions
The answer is
← Didn't Know|Knew It →
Is the following piecewise function continuous for all 
? If not, state where it is discontinuous.
Is the following piecewise function continuous for all
Tap to reveal answer
To check if the piecewise function is continuous, all we need to do is check that the values at 3 and 5 line up.
At 
, this means checking that 
and 
have the same value. More formally, we are checking to see that 
, as to be continuous at a point, a function's left and right limits must both equal the function value at that point.
Plugging 3 into both, we see that both of them are 12 at 
. Thus, they meet up smoothly.
Next, for 
, we have 
and 
. Plugging in 5, we get 22 for both equations.
As all three equations are polynomials, we know they will be continuous everywhere else, and they meet up smoothly at the piecewise bounds, thus ensuring that the function is continuous everywhere.
Note, there are sharp turns at 
and 
, but this only means the function isn't differentiable at these points -- we're only concerned with continuity, which is if the equations meet up. Thus, the function is continuous.
To check if the piecewise function is continuous, all we need to do is check that the values at 3 and 5 line up.
At
Plugging 3 into both, we see that both of them are 12 at
Next, for
As all three equations are polynomials, we know they will be continuous everywhere else, and they meet up smoothly at the piecewise bounds, thus ensuring that the function is continuous everywhere.
Note, there are sharp turns at
← Didn't Know|Knew It →
$f(x)={x^{2}$+3} when and
f(x)={7x-3} when xgeq 3
At the funciton described above is:
$f(x)={x^{2}$+3} when
f(x)={7x-3} when xgeq 3
At
Tap to reveal answer
The answer is both.
If graphed the student will see that the two graphs are continuous at . There is no gap in the graph or no uneven transitions. If the graph is continuous then it is differentiable so it must be both.
The answer is both.
If graphed the student will see that the two graphs are continuous at
← Didn't Know|Knew It →
Which of the following functions contains a removeable discontinuity?
Which of the following functions contains a removeable discontinuity?
Tap to reveal answer
A removeable discontinuity occurs whenever there is a hole in a graph that could be fixed (or "removed") by filling in a single point. Put another way, if there is a removeable discontinuity at , then the limit as approaches exists, but the value of does not.
For example, the function $f(x)=\frac{1+x^3$}{1+x}$ contains a removeable discontinuity at . Notice that we could simplify as follows:
$f(x)=\frac{1+x^3$$}{1+x}$=\frac{(1+x)(x^2$$-x+1)}{1+x}$=x^{2}$-x+1, where xneq -1.
Thus, we could say that $lim_{xrightarrow $-1}$$\frac{1+x^$3$}{1+x}$=lim_{xrightarrow $-1}$x^2$$-x+1=(-1)^2$-(-1)+1=3.
As we can see, the limit of exists at , even though is undefined.
What this means is that will look just like the parabola with the equation $x^{2}$-x+1 EXCEPT when, where there will be a hole in the graph. However, if we were to just define , then we could essentially "remove" this discontinuity. Therefore, we can say that there is a removeable discontinuty at .
The functions
, and
have discontinuities, but these discontinuities occur as vertical asymptotes, not holes, and thus are not considered removeable.
The functions
and $f(x)=\frac{x+1}{1+x^{2}$$} are continuous over all the real values of ; they have no discontinuities of any kind.
The answer is
.
A removeable discontinuity occurs whenever there is a hole in a graph that could be fixed (or "removed") by filling in a single point. Put another way, if there is a removeable discontinuity at
For example, the function $f(x)=\frac{1+x^3$}{1+x}$ contains a removeable discontinuity at
$f(x)=\frac{1+x^3$$}{1+x}$=\frac{(1+x)(x^2$$-x+1)}{1+x}$=x^{2}$-x+1, where xneq -1.
Thus, we could say that $lim_{xrightarrow $-1}$$\frac{1+x^$3$}{1+x}$=lim_{xrightarrow $-1}$x^2$$-x+1=(-1)^2$-(-1)+1=3.
As we can see, the limit of
What this means is that
The functions
have discontinuities, but these discontinuities occur as vertical asymptotes, not holes, and thus are not considered removeable.
The functions
The answer is
← Didn't Know|Knew It →
Is the following piecewise function continuous for all ? If not, state where it is discontinuous.
Is the following piecewise function continuous for all
Tap to reveal answer
To check if the piecewise function is continuous, all we need to do is check that the values at 3 and 5 line up.
At , this means checking that and have the same value. More formally, we are checking to see that , as to be continuous at a point, a function's left and right limits must both equal the function value at that point.
Plugging 3 into both, we see that both of them are 12 at . Thus, they meet up smoothly.
Next, for , we have and . Plugging in 5, we get 22 for both equations.
As all three equations are polynomials, we know they will be continuous everywhere else, and they meet up smoothly at the piecewise bounds, thus ensuring that the function is continuous everywhere.
Note, there are sharp turns at and , but this only means the function isn't differentiable at these points -- we're only concerned with continuity, which is if the equations meet up. Thus, the function is continuous.
To check if the piecewise function is continuous, all we need to do is check that the values at 3 and 5 line up.
At
Plugging 3 into both, we see that both of them are 12 at
Next, for
As all three equations are polynomials, we know they will be continuous everywhere else, and they meet up smoothly at the piecewise bounds, thus ensuring that the function is continuous everywhere.
Note, there are sharp turns at
← Didn't Know|Knew It →
$f(x)={x^{2}$+3} when and
f(x)={7x-3} when xgeq 3
At the funciton described above is:
$f(x)={x^{2}$+3} when
f(x)={7x-3} when xgeq 3
At
Tap to reveal answer
The answer is both.
If graphed the student will see that the two graphs are continuous at . There is no gap in the graph or no uneven transitions. If the graph is continuous then it is differentiable so it must be both.
The answer is both.
If graphed the student will see that the two graphs are continuous at
← Didn't Know|Knew It →
Which of the following functions contains a removeable discontinuity?
Which of the following functions contains a removeable discontinuity?
Tap to reveal answer
A removeable discontinuity occurs whenever there is a hole in a graph that could be fixed (or "removed") by filling in a single point. Put another way, if there is a removeable discontinuity at , then the limit as approaches exists, but the value of does not.
For example, the function $f(x)=\frac{1+x^3$}{1+x}$ contains a removeable discontinuity at . Notice that we could simplify as follows:
$f(x)=\frac{1+x^3$$}{1+x}$=\frac{(1+x)(x^2$$-x+1)}{1+x}$=x^{2}$-x+1, where xneq -1.
Thus, we could say that $lim_{xrightarrow $-1}$$\frac{1+x^$3$}{1+x}$=lim_{xrightarrow $-1}$x^2$$-x+1=(-1)^2$-(-1)+1=3.
As we can see, the limit of exists at , even though is undefined.
What this means is that will look just like the parabola with the equation $x^{2}$-x+1 EXCEPT when, where there will be a hole in the graph. However, if we were to just define , then we could essentially "remove" this discontinuity. Therefore, we can say that there is a removeable discontinuty at .
The functions
, and
have discontinuities, but these discontinuities occur as vertical asymptotes, not holes, and thus are not considered removeable.
The functions
and $f(x)=\frac{x+1}{1+x^{2}$$} are continuous over all the real values of ; they have no discontinuities of any kind.
The answer is
.
A removeable discontinuity occurs whenever there is a hole in a graph that could be fixed (or "removed") by filling in a single point. Put another way, if there is a removeable discontinuity at
For example, the function $f(x)=\frac{1+x^3$}{1+x}$ contains a removeable discontinuity at
$f(x)=\frac{1+x^3$$}{1+x}$=\frac{(1+x)(x^2$$-x+1)}{1+x}$=x^{2}$-x+1, where xneq -1.
Thus, we could say that $lim_{xrightarrow $-1}$$\frac{1+x^$3$}{1+x}$=lim_{xrightarrow $-1}$x^2$$-x+1=(-1)^2$-(-1)+1=3.
As we can see, the limit of
What this means is that
The functions
have discontinuities, but these discontinuities occur as vertical asymptotes, not holes, and thus are not considered removeable.
The functions
The answer is
← Didn't Know|Knew It →
Is the following piecewise function continuous for all ? If not, state where it is discontinuous.
Is the following piecewise function continuous for all
Tap to reveal answer
To check if the piecewise function is continuous, all we need to do is check that the values at 3 and 5 line up.
At , this means checking that and have the same value. More formally, we are checking to see that , as to be continuous at a point, a function's left and right limits must both equal the function value at that point.
Plugging 3 into both, we see that both of them are 12 at . Thus, they meet up smoothly.
Next, for , we have and . Plugging in 5, we get 22 for both equations.
As all three equations are polynomials, we know they will be continuous everywhere else, and they meet up smoothly at the piecewise bounds, thus ensuring that the function is continuous everywhere.
Note, there are sharp turns at and , but this only means the function isn't differentiable at these points -- we're only concerned with continuity, which is if the equations meet up. Thus, the function is continuous.
To check if the piecewise function is continuous, all we need to do is check that the values at 3 and 5 line up.
At
Plugging 3 into both, we see that both of them are 12 at
Next, for
As all three equations are polynomials, we know they will be continuous everywhere else, and they meet up smoothly at the piecewise bounds, thus ensuring that the function is continuous everywhere.
Note, there are sharp turns at
← Didn't Know|Knew It →
$f(x)={x^{2}$+3} when and
f(x)={7x-3} when xgeq 3
At the funciton described above is:
$f(x)={x^{2}$+3} when
f(x)={7x-3} when xgeq 3
At
Tap to reveal answer
The answer is both.
If graphed the student will see that the two graphs are continuous at . There is no gap in the graph or no uneven transitions. If the graph is continuous then it is differentiable so it must be both.
The answer is both.
If graphed the student will see that the two graphs are continuous at
← Didn't Know|Knew It →
Which of the following functions contains a removeable discontinuity?
Which of the following functions contains a removeable discontinuity?
Tap to reveal answer
A removeable discontinuity occurs whenever there is a hole in a graph that could be fixed (or "removed") by filling in a single point. Put another way, if there is a removeable discontinuity at , then the limit as approaches exists, but the value of does not.
For example, the function $f(x)=\frac{1+x^3$}{1+x}$ contains a removeable discontinuity at . Notice that we could simplify as follows:
$f(x)=\frac{1+x^3$$}{1+x}$=\frac{(1+x)(x^2$$-x+1)}{1+x}$=x^{2}$-x+1, where xneq -1.
Thus, we could say that $lim_{xrightarrow $-1}$$\frac{1+x^$3$}{1+x}$=lim_{xrightarrow $-1}$x^2$$-x+1=(-1)^2$-(-1)+1=3.
As we can see, the limit of exists at , even though is undefined.
What this means is that will look just like the parabola with the equation $x^{2}$-x+1 EXCEPT when, where there will be a hole in the graph. However, if we were to just define , then we could essentially "remove" this discontinuity. Therefore, we can say that there is a removeable discontinuty at .
The functions
, and
have discontinuities, but these discontinuities occur as vertical asymptotes, not holes, and thus are not considered removeable.
The functions
and $f(x)=\frac{x+1}{1+x^{2}$$} are continuous over all the real values of ; they have no discontinuities of any kind.
The answer is
.
A removeable discontinuity occurs whenever there is a hole in a graph that could be fixed (or "removed") by filling in a single point. Put another way, if there is a removeable discontinuity at
For example, the function $f(x)=\frac{1+x^3$}{1+x}$ contains a removeable discontinuity at
$f(x)=\frac{1+x^3$$}{1+x}$=\frac{(1+x)(x^2$$-x+1)}{1+x}$=x^{2}$-x+1, where xneq -1.
Thus, we could say that $lim_{xrightarrow $-1}$$\frac{1+x^$3$}{1+x}$=lim_{xrightarrow $-1}$x^2$$-x+1=(-1)^2$-(-1)+1=3.
As we can see, the limit of
What this means is that
The functions
have discontinuities, but these discontinuities occur as vertical asymptotes, not holes, and thus are not considered removeable.
The functions
The answer is
← Didn't Know|Knew It →
Is the following piecewise function continuous for all ? If not, state where it is discontinuous.
Is the following piecewise function continuous for all
Tap to reveal answer
To check if the piecewise function is continuous, all we need to do is check that the values at 3 and 5 line up.
At , this means checking that and have the same value. More formally, we are checking to see that , as to be continuous at a point, a function's left and right limits must both equal the function value at that point.
Plugging 3 into both, we see that both of them are 12 at . Thus, they meet up smoothly.
Next, for , we have and . Plugging in 5, we get 22 for both equations.
As all three equations are polynomials, we know they will be continuous everywhere else, and they meet up smoothly at the piecewise bounds, thus ensuring that the function is continuous everywhere.
Note, there are sharp turns at and , but this only means the function isn't differentiable at these points -- we're only concerned with continuity, which is if the equations meet up. Thus, the function is continuous.
To check if the piecewise function is continuous, all we need to do is check that the values at 3 and 5 line up.
At
Plugging 3 into both, we see that both of them are 12 at
Next, for
As all three equations are polynomials, we know they will be continuous everywhere else, and they meet up smoothly at the piecewise bounds, thus ensuring that the function is continuous everywhere.
Note, there are sharp turns at
← Didn't Know|Knew It →
$f(x)={x^{2}$+3} when and
f(x)={7x-3} when xgeq 3
At the funciton described above is:
$f(x)={x^{2}$+3} when
f(x)={7x-3} when xgeq 3
At
Tap to reveal answer
The answer is both.
If graphed the student will see that the two graphs are continuous at . There is no gap in the graph or no uneven transitions. If the graph is continuous then it is differentiable so it must be both.
The answer is both.
If graphed the student will see that the two graphs are continuous at
← Didn't Know|Knew It →
Which of the following functions contains a removeable discontinuity?
Which of the following functions contains a removeable discontinuity?
Tap to reveal answer
A removeable discontinuity occurs whenever there is a hole in a graph that could be fixed (or "removed") by filling in a single point. Put another way, if there is a removeable discontinuity at , then the limit as approaches exists, but the value of does not.
For example, the function $f(x)=\frac{1+x^3$}{1+x}$ contains a removeable discontinuity at . Notice that we could simplify as follows:
$f(x)=\frac{1+x^3$$}{1+x}$=\frac{(1+x)(x^2$$-x+1)}{1+x}$=x^{2}$-x+1, where xneq -1.
Thus, we could say that $lim_{xrightarrow $-1}$$\frac{1+x^$3$}{1+x}$=lim_{xrightarrow $-1}$x^2$$-x+1=(-1)^2$-(-1)+1=3.
As we can see, the limit of exists at , even though is undefined.
What this means is that will look just like the parabola with the equation $x^{2}$-x+1 EXCEPT when, where there will be a hole in the graph. However, if we were to just define , then we could essentially "remove" this discontinuity. Therefore, we can say that there is a removeable discontinuty at .
The functions
, and
have discontinuities, but these discontinuities occur as vertical asymptotes, not holes, and thus are not considered removeable.
The functions
and $f(x)=\frac{x+1}{1+x^{2}$$} are continuous over all the real values of ; they have no discontinuities of any kind.
The answer is
.
A removeable discontinuity occurs whenever there is a hole in a graph that could be fixed (or "removed") by filling in a single point. Put another way, if there is a removeable discontinuity at
For example, the function $f(x)=\frac{1+x^3$}{1+x}$ contains a removeable discontinuity at
$f(x)=\frac{1+x^3$$}{1+x}$=\frac{(1+x)(x^2$$-x+1)}{1+x}$=x^{2}$-x+1, where xneq -1.
Thus, we could say that $lim_{xrightarrow $-1}$$\frac{1+x^$3$}{1+x}$=lim_{xrightarrow $-1}$x^2$$-x+1=(-1)^2$-(-1)+1=3.
As we can see, the limit of
What this means is that
The functions
have discontinuities, but these discontinuities occur as vertical asymptotes, not holes, and thus are not considered removeable.
The functions
The answer is
← Didn't Know|Knew It →
Is the following piecewise function continuous for all ? If not, state where it is discontinuous.
Is the following piecewise function continuous for all
Tap to reveal answer
To check if the piecewise function is continuous, all we need to do is check that the values at 3 and 5 line up.
At , this means checking that and have the same value. More formally, we are checking to see that , as to be continuous at a point, a function's left and right limits must both equal the function value at that point.
Plugging 3 into both, we see that both of them are 12 at . Thus, they meet up smoothly.
Next, for , we have and . Plugging in 5, we get 22 for both equations.
As all three equations are polynomials, we know they will be continuous everywhere else, and they meet up smoothly at the piecewise bounds, thus ensuring that the function is continuous everywhere.
Note, there are sharp turns at and , but this only means the function isn't differentiable at these points -- we're only concerned with continuity, which is if the equations meet up. Thus, the function is continuous.
To check if the piecewise function is continuous, all we need to do is check that the values at 3 and 5 line up.
At
Plugging 3 into both, we see that both of them are 12 at
Next, for
As all three equations are polynomials, we know they will be continuous everywhere else, and they meet up smoothly at the piecewise bounds, thus ensuring that the function is continuous everywhere.
Note, there are sharp turns at
← Didn't Know|Knew It →
$f(x)={x^{2}$+3} when and
f(x)={7x-3} when xgeq 3
At the funciton described above is:
$f(x)={x^{2}$+3} when
f(x)={7x-3} when xgeq 3
At
Tap to reveal answer
The answer is both.
If graphed the student will see that the two graphs are continuous at . There is no gap in the graph or no uneven transitions. If the graph is continuous then it is differentiable so it must be both.
The answer is both.
If graphed the student will see that the two graphs are continuous at
← Didn't Know|Knew It →
Which of the following functions contains a removeable discontinuity?
Which of the following functions contains a removeable discontinuity?
Tap to reveal answer
A removeable discontinuity occurs whenever there is a hole in a graph that could be fixed (or "removed") by filling in a single point. Put another way, if there is a removeable discontinuity at , then the limit as approaches exists, but the value of does not.
For example, the function $f(x)=\frac{1+x^3$}{1+x}$ contains a removeable discontinuity at . Notice that we could simplify as follows:
$f(x)=\frac{1+x^3$$}{1+x}$=\frac{(1+x)(x^2$$-x+1)}{1+x}$=x^{2}$-x+1, where xneq -1.
Thus, we could say that $lim_{xrightarrow $-1}$$\frac{1+x^$3$}{1+x}$=lim_{xrightarrow $-1}$x^2$$-x+1=(-1)^2$-(-1)+1=3.
As we can see, the limit of exists at , even though is undefined.
What this means is that will look just like the parabola with the equation $x^{2}$-x+1 EXCEPT when, where there will be a hole in the graph. However, if we were to just define , then we could essentially "remove" this discontinuity. Therefore, we can say that there is a removeable discontinuty at .
The functions
, and
have discontinuities, but these discontinuities occur as vertical asymptotes, not holes, and thus are not considered removeable.
The functions
and $f(x)=\frac{x+1}{1+x^{2}$$} are continuous over all the real values of ; they have no discontinuities of any kind.
The answer is
.
A removeable discontinuity occurs whenever there is a hole in a graph that could be fixed (or "removed") by filling in a single point. Put another way, if there is a removeable discontinuity at
For example, the function $f(x)=\frac{1+x^3$}{1+x}$ contains a removeable discontinuity at
$f(x)=\frac{1+x^3$$}{1+x}$=\frac{(1+x)(x^2$$-x+1)}{1+x}$=x^{2}$-x+1, where xneq -1.
Thus, we could say that $lim_{xrightarrow $-1}$$\frac{1+x^$3$}{1+x}$=lim_{xrightarrow $-1}$x^2$$-x+1=(-1)^2$-(-1)+1=3.
As we can see, the limit of
What this means is that
The functions
have discontinuities, but these discontinuities occur as vertical asymptotes, not holes, and thus are not considered removeable.
The functions
The answer is
← Didn't Know|Knew It →
Is the following piecewise function continuous for all ? If not, state where it is discontinuous.
Is the following piecewise function continuous for all
Tap to reveal answer
To check if the piecewise function is continuous, all we need to do is check that the values at 3 and 5 line up.
At , this means checking that and have the same value. More formally, we are checking to see that , as to be continuous at a point, a function's left and right limits must both equal the function value at that point.
Plugging 3 into both, we see that both of them are 12 at . Thus, they meet up smoothly.
Next, for , we have and . Plugging in 5, we get 22 for both equations.
As all three equations are polynomials, we know they will be continuous everywhere else, and they meet up smoothly at the piecewise bounds, thus ensuring that the function is continuous everywhere.
Note, there are sharp turns at and , but this only means the function isn't differentiable at these points -- we're only concerned with continuity, which is if the equations meet up. Thus, the function is continuous.
To check if the piecewise function is continuous, all we need to do is check that the values at 3 and 5 line up.
At
Plugging 3 into both, we see that both of them are 12 at
Next, for
As all three equations are polynomials, we know they will be continuous everywhere else, and they meet up smoothly at the piecewise bounds, thus ensuring that the function is continuous everywhere.
Note, there are sharp turns at
← Didn't Know|Knew It →
$f(x)={x^{2}$+3} when and
f(x)={7x-3} when xgeq 3
At the funciton described above is:
$f(x)={x^{2}$+3} when
f(x)={7x-3} when xgeq 3
At
Tap to reveal answer
The answer is both.
If graphed the student will see that the two graphs are continuous at . There is no gap in the graph or no uneven transitions. If the graph is continuous then it is differentiable so it must be both.
The answer is both.
If graphed the student will see that the two graphs are continuous at
← Didn't Know|Knew It →
Which of the following functions contains a removeable discontinuity?
Which of the following functions contains a removeable discontinuity?
Tap to reveal answer
A removeable discontinuity occurs whenever there is a hole in a graph that could be fixed (or "removed") by filling in a single point. Put another way, if there is a removeable discontinuity at , then the limit as approaches exists, but the value of does not.
For example, the function $f(x)=\frac{1+x^3$}{1+x}$ contains a removeable discontinuity at . Notice that we could simplify as follows:
$f(x)=\frac{1+x^3$$}{1+x}$=\frac{(1+x)(x^2$$-x+1)}{1+x}$=x^{2}$-x+1, where xneq -1.
Thus, we could say that $lim_{xrightarrow $-1}$$\frac{1+x^$3$}{1+x}$=lim_{xrightarrow $-1}$x^2$$-x+1=(-1)^2$-(-1)+1=3.
As we can see, the limit of exists at , even though is undefined.
What this means is that will look just like the parabola with the equation $x^{2}$-x+1 EXCEPT when, where there will be a hole in the graph. However, if we were to just define , then we could essentially "remove" this discontinuity. Therefore, we can say that there is a removeable discontinuty at .
The functions
, and
have discontinuities, but these discontinuities occur as vertical asymptotes, not holes, and thus are not considered removeable.
The functions
and $f(x)=\frac{x+1}{1+x^{2}$$} are continuous over all the real values of ; they have no discontinuities of any kind.
The answer is
.
A removeable discontinuity occurs whenever there is a hole in a graph that could be fixed (or "removed") by filling in a single point. Put another way, if there is a removeable discontinuity at
For example, the function $f(x)=\frac{1+x^3$}{1+x}$ contains a removeable discontinuity at
$f(x)=\frac{1+x^3$$}{1+x}$=\frac{(1+x)(x^2$$-x+1)}{1+x}$=x^{2}$-x+1, where xneq -1.
Thus, we could say that $lim_{xrightarrow $-1}$$\frac{1+x^$3$}{1+x}$=lim_{xrightarrow $-1}$x^2$$-x+1=(-1)^2$-(-1)+1=3.
As we can see, the limit of
What this means is that
The functions
have discontinuities, but these discontinuities occur as vertical asymptotes, not holes, and thus are not considered removeable.
The functions
The answer is
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