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High School Math : Limits

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Example Question #1 : Limits

$f(x)=\frac{x^{2}-x-6}{x+2}$

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

Possible Answers:

Correct answer:

Explanation:

A limit describes what -value a function approaches as approaches a certain value (in this case, ). The easiest way to find what -value a function approaches is to substitute the -value into the equation.

$f(-2)=\frac{(-2)^{2}-(-2)-6}{-2+2}=\frac{4+2-6}{0}=\frac{0}{0}$

Substituting for gives us an undefined value (which is NOT the same thing as 0). This means the function is not defined at that point. However, just because a function is undefined at a point doesn't mean it doesn't have a limit. The limit is simply whichever value the function is getting close to.

One method of finding the limit is to try and simplify the equation as much as possible:

$f(x)=\frac{(x-3)(x+2)}{x+2}$

As you can see, there are common factors between the numerator and the denominator that can be canceled out. (Remember, when you cancel out a factor from a rational equation, it means that the function has a hole -- an undefined point -- where that factor equals zero.)

After canceling out the common factors, we're left with:

f(x)=x-3

Even though the domain of the original function is restricted ( cannot equal ), we can still substitute into this simplified equation to find the limit at x=-2.

f(-2)=-2-3=-5

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Example Question #1 : Finding One Sided Limits

Let $f(x)=\frac{1}{x}$ .

Find $\lim_{x\ \to \0^{-}} \frac{1}{x}$ .

Possible Answers:

$\infty$

The limit does not exist.

$-\infty$

Correct answer:

$-\infty$

Explanation:

1overx

This is a graph of $\frac{1}{x}$ . We know that $\frac{1}{0}$ is undefined; therefore, there is no value for f(0) . But as we take a look at the graph, we can see that as approaches 0 from the left, f(x) approaches negative infinity.

This can be illustrated by thinking of small negative numbers.

f(-.1)=-10

f(-.01)=-100

NOTE: Pay attention to one-sided limit specifications, as it is easy to pick the wrong answer choice if you're not careful.

$\lim_{x \to \0^+} \frac{1}{x}$ is actually infinity, not negative infinity.

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Example Question #3 : Limits

Evaluate the limit below:

$y=\lim_{x \to\frac{\pi }2{} }(sinx-cosx)^{tanx}$

Possible Answers:

{e}

$\frac{1}{e}$

$\frac{\pi }{2}$

Correct answer:

$\frac{1}{e}$

Explanation:

will approach $1^{\infty }$ when approaches $\frac{\pi }{2}$ , so $\lim_{x\rightarrow \frac{\pi }{2}} (lny)$ will be of type $0\cdot \infty$ as shown below:

$\lim_{x\rightarrow \frac{\pi }{2}} lny = \lim_{x\rightarrow \frac{\pi }{2}} ln (sinx-cosx)^{tanx}=\lim_{x\rightarrow \frac{\pi }{2}} tanx\cdot ln(sinx-cosx) = \infty \cdot 0$

So, we can apply the L’ Hospital's Rule:

$\lim_{x\rightarrow \frac{\pi }{2}} lny = \lim_{x\rightarrow \frac{\pi }{2}} \frac{sinx \cdot ln(sinx-cosx)}{cosx}=$

$\lim_{x\rightarrow \frac{\pi }{2}} \frac{cosx\cdot ln(sinx-cosx)+sinx\cdot (\frac{cosx+sinx}{sinx-cosx})}{-sinx}=\frac{0+1}{-1}=-1$

since:

$\lim_{x\rightarrow \frac{\pi }{2}} lny=-1$

hence:

$\lim_{x\rightarrow \frac{\pi }{2}} y=\lim_{x\rightarrow \frac{\pi }{2}} e^{lny}=e^{-1}=\frac{1}{e}$

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Example Question #1 : Finding Limits As X Approaches Infinity

$\textup{Find: }\lim_{x \to \infty }\frac{2x^{2}-7x}{x^{2}+3}$

Possible Answers:

$\infty$

Correct answer:

Explanation:

$\textup{As } x \to\infty,\textup{all but the terms with the highest exponents become insignificant.}$

$\textup{As }x^{2} \textup{ is the highest on top and bottom, evaluate the coefficients:}} \frac{2}{1}=2$

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Example Question #5 : Limits

Calculate $\lim_{x\rightarrow \infty }\left ( x^{2} \sin \frac{1}{ x^{2}-1} - \sin \frac{1}{ x^{2}-1} \right )$ .

Possible Answers:

The limit does not exist.

$\infty$

$\sin 1$

Correct answer:

Explanation:

This can be rewritten as follows:

$\lim_{x\rightarrow \infty }\left ( x^{2} \sin \frac{1}{ x^{2}-1} - \sin \frac{1}{ x^{2}-1} \right )$

$= \lim_{x\rightarrow \infty }\left ( x^{2}\cdot \sin \frac{1}{ x^{2}-1} -1\cdot \sin \frac{1}{ x^{2}-1} \right )$

$= \lim_{x\rightarrow \infty }\left [\left ( x^{2}-1 \right ) \cdot \sin \frac{1}{ x^{2}-1} \right]$

$= \lim_{x\rightarrow \infty } \frac{ \sin \frac{1}{ x^{2}-1} }{\frac{1}{ x^{2}-1}}$

We can substitute $u = \frac{1}{ x^{2}-1}$ , noting that as $x\rightarrow \infty$ , $u \rightarrow 0$ :

$= \lim_{u\rightarrow 0} \frac{ \sin u }{u} = 1$ , which is the correct choice.

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Example Question #2071 : High School Math

The speed of a car traveling on the highway is given by the following function of time:

v(t) = 5t^2+2t

What can you say about the car's speed after a long time (that is, as approaches infinity)?

Possible Answers:

The speed of the car approaches infinity.

The speed of the car approaches zero.

The speed of the car approaches a constant number.

Nothing can be concluded from the given function.

The speed of the car depends on the starting speed.

Correct answer:

The speed of the car approaches infinity.

Explanation:

The function given is a polynomial with a term t^n , such that is greater than 1.

Whenever this is the case, we can say that the whole function diverges (approaches infinity) in the limit as approaches infinity.

This tells us that the given function is not a very realistic description of a car's speed for large !

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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 : Limits

Example Question #1 : Finding One Sided Limits

Example Question #3 : Limits

Example Question #1 : Finding Limits As X Approaches Infinity

Example Question #5 : Limits

Example Question #2071 : High School Math

All High School Math Resources

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