AP Calculus AB - Calculating limits using algebra

Define $f (x) = \begin{Bmatrix} x^{2}+2x+ 3& x< 0 \ a \cos\theta x & x \ge 0 \end{matrix}$ for some real $a,\theta$ .

Evaluate and $\theta$ so that is both continuous and differentiable at .

No such values exist.

$a = 3, \theta = 1$

$a = 3, \theta = 2$

$a = 1, \theta = 3$

$a = 2, \theta = 3$

Explanation

For to be continuous, it must hold that

$\lim_{x \rightarrow 0^{-}} f(x) = \lim_{x \rightarrow 0^{+}} f(x)$ .

To find $\lim_{x \rightarrow 0^{-}} f(x)$ , we can use the definition of for all negative values of :

$\lim_{x \rightarrow 0^{-}} f(x) =\lim_{x \rightarrow 0^{-}} ( x^{2}+2x+ 3) = 0^{2}+ 2(0)+ 3 = 3$

It must hold that $\lim_{x \rightarrow 0^{+}} f(x) =3$ as well; using the definition of for all positive values of :

$\lim_{x \rightarrow 0^{+}} f(x) =\lim_{x \rightarrow 0^{+}} a \cos \theta x = a \cos 0 = a \cdot 1 = a$

$\lim_{x \rightarrow 0^{+}} f(x) =3$ , so .

Now examine . For to be differentiable, it must hold that

$\lim_{x \rightarrow 0^{-}} f'(x) = \lim_{x \rightarrow 0^{+}} f'(x)$ .

To find $\lim_{x \rightarrow 0^{-}} f'(x)$ , we can differentiate the expression for for all negative values of :

$f(x)= x^{2}+ 2x + 3$

$\lim_{x \rightarrow 0^{-}} f'(x) = \lim_{x \rightarrow 0^{-}} (2x +2)= 2(0)+2 = 2$

To find $\lim_{x \rightarrow 0^{+}} f'(x)$ , we can differentiate the expression for for all positive values of :

$f(x) = a \cos \theta x$

$f'(x) = - a \theta \sin \theta x$

$\lim_{x \rightarrow 0^{+}} f'(x) =\lim_{x \rightarrow 0^{+}} ( - a \theta \sin \theta x)$

We know that , so

$\lim_{x \rightarrow 0^{+}} f'(x) =\lim_{x \rightarrow 0^{+}} ( - 3 \theta \sin \theta x) = -3 \theta \sin 0 = -3 \theta \cdot 0 = 0$

Since $\lim_{x \rightarrow 0^{-}} f'(x) = 2$ and $\lim_{x \rightarrow 0^{+}} f'(x)=0$ ,

$\lim_{x \rightarrow 0 } f'(x)$ cannot exist regardless of the values of and $\theta$ .

Evaluate the following limit:

$\lim_{x\rightarrow 4}\frac{x^{2}+2x-24}{x-4}$

Explanation

Factor x-4 out of the numerator and simplify:

$\frac{(x-4)(x+6)}{(x-4)}=x+6$

Evaluate the limit for x=4:

Although there is a discontinuity at x=4, the limit at x=4 is 10 because the function approaches ten from the left and right side.

$f(x) = \left{\begin{matrix} \frac{2}{x+2}&x<0 \ 0 &x= 0 \ 2x+1 & x > 0 \end{matrix}\right.$

Which of the following is equal to $\lim_{x\rightarrow 0 }f(x)$ ?

$\frac{1}{2}$

$\lim_{x\rightarrow 0 }f(x)$ does not exist, because $\lim_{x\rightarrow 0 - }f(x) e \lim_{x\rightarrow 0 + }f(x)$ .

$\lim_{x\rightarrow 0 }f(x)$ does not exist, because either or both of $\lim_{x\rightarrow 0 - }f(x)$ and $\lim_{x\rightarrow 0 + }f(x)$ is unequal to .

Explanation

The limit of a function as approaches a value $x_{0}$ exists if and only if the limit from the left is equal to the limit from the right; the actual value of $f(x_{0})$ is irrelevant. Since the function is piecewise-defined, we can determine whether these limits are equal by finding the limits of the individual expressions. These are both polynomials, so the limits can be calculated using straightforward substitution:

$\lim_{x\rightarrow 0 - }f(x)= \lim_{x\rightarrow 0 - } \frac{2}{x+2 } = \frac{2}{0+2 } = 1$

$\lim_{x\rightarrow 0 + }f(x)= \lim_{x\rightarrow 0 - } 2x+1 = 2(0)+1 = 1$

$\lim_{x\rightarrow 0 - }f(x) = \lim_{x\rightarrow 0 + }f(x) = 1$ , so $\lim_{x\rightarrow 0 }f(x) = 1$ .

Evaluate the following limit:

$\lim_{x\rightarrow 0}\frac{x^3+5x^2+8x}{x}$

does not exist

Explanation

Factor the numerator to evaluate the limit:

$\frac{x(x+2)(x+4)}{x}=(x+2)(x+4)$

Evaluate the limit:

$\lim_{x\rightarrow 0}(x+2)(x+4)=(2)(4)=8$

There is a discontinuity at x=0 but the limit is equal to 8 because the limit from the right is equal to the limit from the left.

Calculate $\lim_{x\rightarrow 5} \frac{x^2-25}{x-5}$

The limit does not exist

$\infty$

Explanation

First we notice that substituting 5 in for x will give us a 0 in the denominator.

So we simplify the equation by noticing the numerator is the difference of two squares.

$\frac{x^2-25}{x-5}=\frac{(x+5)(x-5)}{x-5}=x+5$

Now we can substitute 5 in for x, and we arrive at our answer of 10.

$\lim_{x\rightarrow 2}\frac{x^2+5x-14}{x-2}=?$

$\infty$

Undefined

$-\infty$

Explanation

To start this problem, we need to factor the quadratic equation. This will result in a hole at , which will allow us to directly substitute the limit value into the resulting function to determine the limit.

$\frac{x^2+5x-14}{x-2}=\frac{(x-2)(x+7)}{x-2}=x+7.$

$\lim_{x\rightarrow 2}{x+7}=2+7=9.$

Find the limit:

$\lim_{x\rightarrow \propto }\frac{4x^3+3x^2+10}{10x^3+x^2+8x}$

$\frac{4}{10}$

$\propto$

DNE (Does not exist)

Explanation

So for limits involving infinity, there is one important concept regarding fractions that is important to understand.

So if you have a fraction with a number on the bottom that is getting larger and larger, the whole fraction becomes smaller.

For example $\frac{1}{1000}>\frac{1}{10000}$

So the way to solve for limits involving infinity is to divide each term on the top, and each term on the bottom by the largest power of the variable. In the case of this problem, that is .

So if you do this you get:

$\lim_{x\rightarrow \propto }\frac{\frac{4x^3}{x^3}+\frac{3x^2}{x^3}+\frac{10}{x^3}}{\frac{10x^3}{x^3}+\frac{x^2}{x^3}+\frac{8x}{x^3}}$

So anything divided by itself is 1, which means that the first two terms cancel to one. Then the rest of the terms with a higher power of x on the bottom than on the top will have some power of x left on the bottom. This will look like:

$\lim_{x\rightarrow \propto }\frac{4+\frac{3}{x}+\frac{10}{x^3}}{10+\frac{1}{x}+\frac{8}{x^2}}$

Then, if you let the limit go to infinity, the terms with an x left on the bottom will go to zero.

This leaves you with the answer of $\frac{4}{10}$

$\lim_{x\rightarrow -3}\frac{2x+4}{x-3}=?$

$\frac{1}{3}$

$\frac{-1}{6}$

Undefined.

$\frac{1}{6}$

$\frac{-1}{3}$

Explanation

To evaluate this limit, we need to direct substitute the value in question into our function. Note, this value is in the domain of the function, so direct substitution is the only method we need here.

$\lim_{x\rightarrow -3}\frac{2x+4}{x-3}=\frac{2(-3)+4}{(-3-3)}=\frac{-2}{-6}=\frac{1}{3}.$

$\begin{align*}&\text{Find the limit of }f(x)=\frac{4x^{2} - 68x - 152}{2x^{2} - 62x + 456}\&\text{At the location }x=19\end{align*}$

$\frac{1}{6}$

$-\frac{1}{6}$

Explanation

$\begin{align*}&\text{Simply plugging in the value of }x=19\text{ into the function}\&\frac{4x^{2} - 68x - 152}{2x^{2} - 62x + 456}\&\text{will not tell us much, because it simply gives us a zero in}\&\text{the numerator and denominator: }\frac{0}{0}\&\text{Note, however, that we're calculating a limit. As such, we can try to simplify our fraction.}\&\text{We can do this because we're not actually evaluating the function at these zero points,}\&\text{just close to them. Factoring our function we find:}\&\frac{4\cdot (x + 2)\cdot (x - 19)}{2\cdot (x - 12)\cdot (x - 19)}\&\text{Which reduces to:}\&\frac{(4x + 8)}{(2x - 24)}\&\text{At the value of }x=19\text{ the function is}\&6\&\text{*Note*: To find roots, you can always utilize the quadratic formula:}\&\text{For equations of the form: }ax^2+bx+c\&\text{The roots are: }x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\end{align*}$

$\begin{align*}&\text{Find the limit of }f(x)=\frac{5x^{2} - 41x - 36}{x^{2} - 21x + 108}\&\text{At the location }x=9\end{align*}$

$-\frac{49}{3}$

$\frac{49}{3}$

$-\frac{3}{49}$

$\frac{3}{49}$

Explanation

$\begin{align*}&\text{Simply plugging in the value of }x=9\text{ into the function}\&\frac{5x^{2} - 41x - 36}{x^{2} - 21x + 108}\&\text{will not tell us much, because it simply gives us a zero in}\&\text{the numerator and denominator: }\frac{0}{0}\&\text{Note, however, that we're calculating a limit. As such, we can try to simplify our fraction.}\&\text{We can do this because we're not actually evaluating the function at these zero points,}\&\text{just close to them. Factoring our function we find:}\&\frac{(5x + 4)\cdot (x - 9)}{(x - 9)\cdot (x - 12)}\&\text{Which reduces to:}\&\frac{(5x + 4)}{(x - 12)}\&\text{At the value of }x=9\text{ the function is}\&-\frac{49}{3}\&\text{*Note*: To find roots, you can always utilize the quadratic formula:}\&\text{For equations of the form: }ax^2+bx+c\&\text{The roots are: }x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\end{align*}$

Calculating limits using algebra

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