Defining Limits and Using Limit Notation
Help Questions
AP Calculus AB › Defining Limits and Using Limit Notation
For $x\ne0$, $s(x)=\dfrac{|x|}{x}$ and $s(0)=0$. Which expression represents the right-hand limit as $x$ approaches $0$?
$\lim_{x\to0^-} s(x)=1$
$\lim_{x\to0^+} s(x)=-1$
$\lim_{x\to0} s(0)=0$
$\lim_{x\to0} s(x)=0$
$\lim_{x\to0^+} s(x)=1$
Explanation
The function $s(x) = \frac{|x|}{x}$ equals $\frac{x}{x} = 1$ when $x>0$ and $\frac{-x}{x} = -1$ when $x<0$. For the right-hand limit as $x\to 0^+$, we consider positive values of $x$ approaching 0, where $s(x) = 1$. Therefore, $\lim_{x\to 0^+} s(x) = 1$ is the correct notation. The assigned value $s(0)=0$ is irrelevant to the limit calculation—limits describe behavior near a point, not at it. A common error is writing $\lim_{x\to 0^+} s(0)$, which incorrectly substitutes 0 into the function within the limit notation. Note that the left-hand limit would be -1, so the two-sided limit doesn't exist. Notation checklist: (1) $\lim$ symbol, (2) $x\to 0^+$ for right approach, (3) $s(x)$ not $s(0)$, (4) equals 1.
For the function $f$ shown in the table, which limit expression represents the behavior of $f(x)$ as $x$ approaches $2$?
$\displaystyle \lim_{x\to 2^-} f(x)=3$
$\displaystyle \lim_{x\to 2} f(x)=5$
$\displaystyle \lim_{x\to 5} f(x)=2$
$\displaystyle \lim_{x\to 2} f(2)=5$
$\displaystyle f(2)=5$
Explanation
Limit notation is used to describe the value a function approaches as the input nears a certain point, distinct from the function's value at that point. In this scenario, the table indicates that f(x) approaches 5 as x gets close to 2 from both sides, making $\lim_{x\to 2}$ f(x)=5 the appropriate expression. This notation is valid because it focuses on the behavior around x=2 without evaluating f at exactly 2. A common symbolic error is writing $\lim_{x\to 2}$ f(2)=5, which improperly substitutes the point into the function inside the limit. Another frequent mistake is using f(2)=5, which represents the function value, not the limit. Remember, limits can exist even if the function is undefined at the point. Transferable notation checklist: 1. Use 'lim' to denote limits. 2. Specify the approach with x\to a. 3. Add ^+ or ^- for one-sided limits if needed. 4. Ensure the expression equals the approached value.
In the table, $t(x)$ approaches $6$ as $x$ approaches $1$. Which limit expression correctly represents this behavior?
$\displaystyle \lim_{x\to 6} t(x)=1$
$\displaystyle t(1)=6$
$\displaystyle \lim_{x\to 1^-} t(x)=5$
$\displaystyle \lim_{x\to 1} t(x)=6$
$\displaystyle \lim_{x\to 1} t(1)=6$
Explanation
Limit notation conveys approaching values, as the table shows t(x) nearing 6 as x approaches 1. Hence, $\lim_{x\to 1} t(x)=6$ is appropriate for this behavior. This is valid regardless of $t(1)$. An error is $\lim_{x\to 1} t(1)=6$, misplacing evaluation. $t(1)=6$ is function value, not limit. Distinguish clearly. Transferable notation checklist: 1. Use 'lim' for limits. 2. Specify the approach with $x \to a$. 3. Add $^{+}$ or $^{-}$ for one-sided limits if needed. 4. Ensure the expression equals the approached value.
The table indicates $q(x)$ approaches $0$ as $x$ approaches $-2$. Which limit expression matches this behavior?
$\displaystyle \lim_{x\to -2} q(x)=0$
$\displaystyle \lim_{x\to -2^+} q(x)=2$
$\displaystyle \lim_{x\to 0} q(x)=-2$
$\displaystyle \lim_{x\to -2} q(-2)=0$
$\displaystyle q(-2)=0$
Explanation
Limit notation describes functional behavior near a point, with the table indicating q(x) approaches 0 as x nears -2. Thus, $\lim_{x\to -2}$ q(x)=0 represents this two-sided approach. This is valid as it ignores q(-2) and focuses on vicinity. A common error is $\lim_{x\to -2}$ q(-2)=0, blending notation improperly. q(-2)=0 denotes value at -2, not limit. Limits exist independently of point values. Transferable notation checklist: 1. Use 'lim' for limits. 2. Specify the approach with x\to a. 3. Add ^+ or ^- for one-sided limits if needed. 4. Ensure the expression equals the approached value.
The table suggests $c(x)$ approaches $-1$ as $x$ approaches $4$. Which limit expression correctly represents this behavior?
$\displaystyle \lim_{x\to 4^-} c(x)=1$
$\displaystyle \lim_{x\to 4} c(x)=-1$
$\displaystyle c(4)=-1$
$\displaystyle \lim_{x\to 4} c(4)=-1$
$\displaystyle \lim_{x\to -1} c(x)=4$
Explanation
Table suggests c(x) to -1 at x=4. $\lim_{x\to 4}$ c(x)=-1 represents. Valid both sides. Error: $\lim_{x\to 4}$ c(4)=-1. c(4)=-1 value. Separate. Transferable notation checklist: 1. Use 'lim' for limits. 2. Specify the approach with x\to a. 3. Add ^+ or ^- for one-sided limits if needed. 4. Ensure the expression equals the approached value.
For $s(x)=\frac{|x|}{x}$ when $x\ne0$ and $s(0)=0$, which limit notation describes $s(x)$ as $x\to0$?
$\displaystyle \lim_{x\to 0^-} s(x)=1$
$\displaystyle \lim_{x\to 0^+} s(x)=-1$
$\displaystyle \lim_{x\to 0} s(x)=0$
$\displaystyle \lim_{x\to 0} s(x)=\text{DNE}$
$\displaystyle \lim_{s(x)\to 0} x=1$
Explanation
If left and right limits differ, the two-sided limit does not exist, denoted as DNE in notation. For s(x) = |x|/x, it approaches 1 from the right and -1 from the left as x nears 0, so $\lim_{x \to 0}$ s(x) = DNE is accurate. This is valid when sides disagree, despite s(0) = 0. A common error is claiming a limit value like 0, confusing with the function at 0. Another symbolic mistake is reversing, such as $\lim_{s(x) \to 0}$ x = 1, or using one-sided without specifying DNE for two-sided. Always check both directions for existence. Transferable notation checklist: 1. Write as $\lim_{x \to a}$ f(x) = L, with x approaching a and f(x) to L. 2. Use + or - for one-sided limits if specified. 3. Do not equate limit to f(a) unless continuous. 4. Avoid swapping x and f(x) roles. 5. Confirm left and right agreement for two-sided limits.
A table indicates $h(x)$ approaches $-2$ as $x$ approaches $0$ from the right; which notation matches?
$\displaystyle \lim_{x\to 0} h(x)=h(0)$
$\displaystyle \lim_{x\to 0^+} h(x)=-2$
$\displaystyle \lim_{x\to -2} h(x)=0$
$\displaystyle \lim_{x\to 0} h(x)=2$
$\displaystyle \lim_{x\to 0^-} h(x)=-2$
Explanation
One-sided limit notation specifies direction using + for right or - for left, which is crucial when behavior differs on each side. The table shows h(x) approaching -2 only from the right as x nears 0, so $\lim_{x \to 0^+}$ h(x) = -2 accurately represents this. This notation is valid as it matches the directional approach described. A common symbolic error is omitting the direction, like $\lim_{x \to 0}$ h(x) = -2, assuming two-sided without evidence. Another mistake is confusing the limit with function equality, such as $\lim_{x \to 0}$ h(x) = h(0). Always check if the data specifies one side or both. Transferable notation checklist: 1. Write as $\lim_{x \to a}$ f(x) = L, with x approaching a and f(x) to L. 2. Use + or - for one-sided limits if specified. 3. Do not equate limit to f(a) unless continuous. 4. Avoid swapping x and f(x) roles. 5. Confirm left and right agreement for two-sided limits.
For $g(x)=\frac{x^2-9}{x-3}$ when $x\ne3$ and $g(3)=10$, which expression represents the limit as $x\to3$?
$\displaystyle \lim_{x\to 10} g(x)=3$
$\displaystyle \lim_{x\to 3} g(x)=10$
$\displaystyle \lim_{x\to 3^-} g(x)=10$
$\displaystyle \lim_{x\to 3} g(x)=6$
$\displaystyle \lim_{g(x)\to 3} x=6$
Explanation
Limit notation for rational functions often involves simplifying expressions to find the approaching value, ignoring the function's defined value at the point. For g(x) = $(x^2$ - 9)/(x - 3) simplified to x + 3 for x ≠ 3, the limit as x approaches 3 is 6, despite g(3) = 10, so $\lim_{x \to 3}$ g(x) = 6 is correct. This is valid because limits consider behavior near the point, not at it. A common error is using the redefined value, like $\lim_{x \to 3}$ g(x) = 10. Another mistake is reversing the limit, such as $\lim_{x \to 10}$ g(x) = 3 or $\lim_{g(x) \to 3}$ x = 6. The two-sided limit applies here as the function approaches the same value from both sides. Transferable notation checklist: 1. Write as $\lim_{x \to a}$ f(x) = L, with x approaching a and f(x) to L. 2. Use + or - for one-sided limits if specified. 3. Do not equate limit to f(a) unless continuous. 4. Avoid swapping x and f(x) roles. 5. Confirm left and right agreement for two-sided limits.
A function $g$ satisfies $g(x)=1$ for $x<3$, $g(3)=5$, and $g(x)=4$ for $x>3$. Which limit notation matches $g$ near $x=3$?
$\lim_{x\to3^+} g(x)=1$
$\lim_{x\to3} g(3)=5$
$\lim_{x\to3^-} g(x)=4$
$\lim_{x\to3^+} g(x)=4$
$\lim_{x\to3} g(x)=5$
Explanation
The function $g$ has a jump discontinuity at $x=3$: it equals 1 for $x<3$, jumps to 5 at $x=3$, then equals 4 for $x>3$. When approaching from the right (values greater than 3), we're in the region where $g(x)=4$, so the right-hand limit is $\lim_{x\to 3^+} g(x) = 4$. The superscript plus sign indicates we only consider values approaching from the right. A common mistake is thinking the limit must involve the function value at the point—but $g(3)=5$ is irrelevant to the right-hand limit. Another error is writing $\lim_{x\to 3} g(3)$, which incorrectly substitutes the value into the limit notation. Notation checklist: (1) $\lim$ symbol, (2) $x\to 3^+$ for right-hand approach, (3) $g(x)$ not $g(3)$, (4) equals 4.
A function $r$ satisfies $r(x)=2$ for $x\le4$ and $r(x)=2+(x-4)$ for $x>4$. Which limit expression matches $r(x)$ as $x\to4^+$?
$\lim_{x\to4} r(x)=4$
$\lim_{x\to4^+} r(4)=2$
$\lim_{x\to4^-} r(x)=3$
$r(4)=3$
$\lim_{x\to4^+} r(x)=2$
Explanation
The function $r$ is defined as $r(x)=2$ for $x\leq 4$ and $r(x)=2+(x-4)$ for $x>4$. For the right-hand limit as $x\to 4^+$, we use the formula for $x>4$: as $x$ approaches 4 from the right, $(x-4)$ approaches 0, so $r(x)$ approaches $2+0=2$. The correct notation is $\lim_{x\to 4^+} r(x) = 2$, where the superscript plus indicates we only consider values greater than 4. Notice that both one-sided limits equal 2, making the function continuous at $x=4$. A common error is writing $\lim_{x\to 4^+} r(4)$, which incorrectly evaluates the function at 4 rather than describing the limiting behavior. Notation checklist: (1) $\lim$ symbol, (2) $x\to 4^+$ for right approach, (3) $r(x)$ not $r(4)$, (4) equals 2.