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

$\infty$

Explanation

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

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

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

$\infty$

Explanation

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

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

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

$\sin 1$

$\infty$

The limit does not exist.

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.

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

$\sin 1$

$\infty$

The limit does not exist.

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.

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

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

The speed of the car approaches infinity.

The speed of the car approaches zero.

The speed of the car approaches a constant number.

The speed of the car depends on the starting speed.

Nothing can be concluded from the given function.

Explanation

The function given is a polynomial with a term , 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 !

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

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

The speed of the car approaches infinity.

The speed of the car approaches zero.

The speed of the car approaches a constant number.

The speed of the car depends on the starting speed.

Nothing can be concluded from the given function.

Explanation

The function given is a polynomial with a term , 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 !

Evaluate the limit below:

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

$\frac{1}{e}$

$\frac{\pi }{2}$

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:

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

Evaluate the limit below:

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

$\frac{1}{e}$

$\frac{\pi }{2}$

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:

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

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

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

$-\infty$

The limit does not exist.

$\infty$

Explanation

1overx

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