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High School Math : Limits

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Example Questions

Example Question #1 : Calculus Ii — Integrals

Function

The above graph depicts a function h(x) . Does $\lim_{x\rightarrow 1} h(x)$ exist, and why or why not?

Possible Answers:

$\lim_{x\rightarrow 1} h(x)$ does not exist because $\lim_{x\rightarrow 1^{-}} h(x) \neq \lim_{x\rightarrow 1^{+}} h(x)$ .

$\lim_{x\rightarrow 1} h(x)$ exists because $f (1 ) = \frac{\lim_{x\rightarrow 1^{-}} h(x) + \lim_{x\rightarrow 1^{+ }} h(x)}{2}$

$\lim_{x\rightarrow 1} h(x)$ does not exist because $f (1 ) \neq \lim_{x\rightarrow 1^{-}} h(x)$

$\lim_{x\rightarrow 1} h(x)$ exists because $\lim_{x\rightarrow 1^{-}} | h(x)| = \lim_{x\rightarrow 1^{+}} |h(x)|$

$\lim_{x\rightarrow 1} h(x)$ does not exist because $f (1 ) \neq \lim_{x\rightarrow 1^{+}} h(x)$

Correct answer:

$\lim_{x\rightarrow 1} h(x)$ does not exist because $\lim_{x\rightarrow 1^{-}} h(x) \neq \lim_{x\rightarrow 1^{+}} h(x)$ .

Explanation:

$\lim_{x\rightarrow 1} h(x)$ exists if and only if $\lim_{x\rightarrow 1^{-}} h(x) = \lim_{x\rightarrow 1^{+}} h(x)$ . As can be seen from the diagram, $\lim_{x\rightarrow 1^{-}} h(x) = 1$ , but $\lim_{x\rightarrow 1^{+}} h(x) = -1$ . Since $\lim_{x\rightarrow 1^{-}} h(x) \neq \lim_{x\rightarrow 1^{+}} h(x)$ , $\lim_{x\rightarrow 1} h(x)$ does not exist.

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Example Question #2 : Calculus Ii — Integrals

Function

The above graph depicts a function f(x) . Does $\lim_{x\rightarrow 1} f(x)$ exist, and why or why not?

Possible Answers:

$\lim_{x\rightarrow 1} f(x)$ does not exist because $f(1) \neq \lim_{x\rightarrow 1^{-}} f(x)$

$\lim_{x\rightarrow 1} f(x)$ does not exist because $\lim_{x\rightarrow 1^{+}} f(x) \neq \lim_{x\rightarrow 1^{-}} f(x)$

$\lim_{x\rightarrow 1} f(x)$ does not exist because $f(1) \neq \lim_{x\rightarrow 1^{+}} f(x)$

$\lim_{x\rightarrow 1} f(x)$ exists because $\lim_{x\rightarrow 1^{+}} f(x) = \lim_{x\rightarrow 1^{-}} f(x)$

$\lim_{x\rightarrow 1} f(x)$ does not exist because is not continuaous at x = 1 .

Correct answer:

$\lim_{x\rightarrow 1} f(x)$ exists because $\lim_{x\rightarrow 1^{+}} f(x) = \lim_{x\rightarrow 1^{-}} f(x)$

Explanation:

$\lim_{x\rightarrow 1} f(x)$ exists if and only if $\lim_{x\rightarrow 1^{+}} f(x) = \lim_{x\rightarrow 1^{-}} f(x)$ ;

the actual value of f(1) is irrelevant, as is whether is continuous there.

As can be seen,

$\lim_{x\rightarrow 1^{+}} f(x) = 1$ and $\lim_{x\rightarrow 1^{-}} f(x) = 1$ ;

therefore, $\lim_{x\rightarrow 1^{+}} f(x) = \lim_{x\rightarrow 1^{-}} f(x)$ ,

and $\lim_{x\rightarrow 1} f(x)$ exists.

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Example Question #3 : Calculus Ii — Integrals

A function is defined by the following piecewise equation:

$f(x)=\left\{\begin{matrix} x^{2}+4x-3, x<3\\10x-12, x\geq 3 \end{matrix}\right.$

At x=3 , the function is:

Possible Answers:

continuous

discontinuous

Correct answer:

continuous

Explanation:

The first step to determine continuity at a point is to determine if the function is defined at that point. When we substitute in 3 for , we get 18 as our -value. x=3 is thus defined for this function.

The next step is determine if the limit of the function is defined at that point. This means that the left-hand limit must be equal to the right-hand limit at x=3 . Substitution reveals the following:

$\lim_{x\rightarrow 3^{-}}f(x)=(3)^{2}+4(3)-3=9+12-3=18$

$\lim_{x\rightarrow 3^{+}}f(x)=10(3)-12=18$

Both sides of the function, therefore, approach a -value of 18.

Finally, we must ensure that the curve is smooth by checking the limit of the derivative of both sides.

$\lim_{x\rightarrow 3^{-}}f^{'}(x)=2(3)+4=10$

$\lim_{x\rightarrow 3^{+}}f^{'}(x)=10$

Since the function passes all three tests, it is continuous.

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Example Question #1 : Limits

Function

The graph depicts a function g(x) . Does $\lim_{x\rightarrow 1} g(x)$ exist?

Possible Answers:

$\lim_{x\rightarrow 1} g(x)$ exists because g(x) is constant on $(-\infty , 1)$ .

$\lim_{x\rightarrow 1} g(x)$ does not exist because g(1) is undefined.

$\lim_{x\rightarrow 1} g(x)$ does not exist because $\lim_{x\rightarrow 1^{-}} g(x) \neq \lim_{x\rightarrow 1^{+}} g(x)$ .

$\lim_{x\rightarrow 1} g(x)$ exists because $\lim_{x\rightarrow 1^{-}} g(x) = \lim_{x\rightarrow 1^{+}} g(x)$ .

$\lim_{x\rightarrow 1} g(x)$ does not exist because is not continuous at x = 1 .

Correct answer:

$\lim_{x\rightarrow 1} g(x)$ exists because $\lim_{x\rightarrow 1^{-}} g(x) = \lim_{x\rightarrow 1^{+}} g(x)$ .

Explanation:

$\lim_{x\rightarrow 1}g(x)$ exists if and only if $\lim_{x\rightarrow 1^{+}} g(x) = \lim_{x\rightarrow 1^{-}} g(x)$ ; the actual value of f(1) is irrelevant.

As can be seen, $\lim_{x\rightarrow 1^{+}} g(x) = 1$ and $\lim_{x\rightarrow 1^{-}}g(x) = 1$ ; therefore, $\lim_{x\rightarrow 1^{+}} g(x) = \lim_{x\rightarrow 1^{-}} g(x)$ , and $\lim_{x\rightarrow 1} g(x)$ exists.

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High School Math : Limits

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #1 : Calculus Ii — Integrals

Example Question #2 : Calculus Ii — Integrals

Example Question #3 : Calculus Ii — Integrals

Example Question #1 : Limits

All High School Math Resources

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