Pre-Calculus - Limits | Practice Hub

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.

Find the limit as x approaches infinity

$\lim_{x\rightarrow \infty }\frac{3x^3-5x^4+4}{14x^2+32x^3}$

$-\infty$

$\infty$

$\frac{3}{32}$

$\frac{-5}{32}$

Explanation

As x approaches infinity we only need to look at the highest order of polynomial in both the numerator and denominator. Then we compare the highest order polynomial in both the numerator and denominator. If the denominator is higher order our limit goes to zero, if the numerator is higher our order our limit goes to positive or negative infinity (depending on the sign of the highest order x term). If our numerator and denominator have the same order the limit goes to a/b where a is the coefficient for the highest order x in the numerator and b is the coefficient for the highest order x in the denominator.

Our numerator has higher order and the coefficient for the x to the fourth term is negative so our limit goes to negative infinity.

Find the limit

$\lim_{x\rightarrow \frac{3}{2}}\frac{6x^2-5x-6}{2x-3}$

$\frac{13}{2}$

$\infty$

$\frac{3}{2}$

$\frac{9}{2}$

Explanation

When x=3/2 our denominator is zero so we can't just plug in 3/2 to get our limit. If we look at the numerator when x=3/2 we find that it is zero as well so our numerator can be factored. We see that our limit can be re-written as:

$\lim_{x\rightarrow \frac{3}{2}}\frac{(2x-3)(3x+2)}{2x-3}$

we then can cancel the 2x-3 from the numerator and denominator leaving us with:

$\lim_{x\rightarrow \frac{3}{2}}3x+2$

and we can just plug in 3/2 into this limit to get

$\frac{13}{2}$

note: our function is not continuous at x=3/2 but the limit does exist.

Find the limit as x approaches infinity

$\lim_{x\rightarrow \infty }\frac{3x^3-5x^4+4}{14x^2+32x^3}$

$-\infty$

$\infty$

$\frac{3}{32}$

$\frac{-5}{32}$

Explanation

Our numerator has higher order and the coefficient for the x to the fourth term is negative so our limit goes to negative infinity.

Find the limit

$\lim_{x\rightarrow \frac{3}{2}}\frac{6x^2-5x-6}{2x-3}$

$\frac{13}{2}$

$\infty$

$\frac{3}{2}$

$\frac{9}{2}$

Explanation

$\lim_{x\rightarrow \frac{3}{2}}\frac{(2x-3)(3x+2)}{2x-3}$

we then can cancel the 2x-3 from the numerator and denominator leaving us with:

$\lim_{x\rightarrow \frac{3}{2}}3x+2$

and we can just plug in 3/2 into this limit to get

$\frac{13}{2}$

note: our function is not continuous at x=3/2 but the limit does exist.

Solve the following limit:

$\lim_{h\rightarrow 0}\frac{(x+h)^2-x^2}{h}$

$\infty$

Explanation

To solve this problem we need to expand the term in the numerator

when we do that we get

$\lim_{h\rightarrow 0}\frac{x^2+2xh+h^2-x^2}{h}$

the second degree x terms cancel and we get

$\lim_{h\rightarrow0}\frac{2xh+h^2}{h}$

now we can cancel our h's in the numerator and denominator to get

$\lim_{h\rightarrow 0}2x+h$

then we can just plug 0 in for h and we get our answer

Solve the following limit:

$\lim_{h\rightarrow 0}\frac{(x+h)^2-x^2}{h}$

$\infty$

Explanation

To solve this problem we need to expand the term in the numerator

when we do that we get

$\lim_{h\rightarrow 0}\frac{x^2+2xh+h^2-x^2}{h}$

the second degree x terms cancel and we get

$\lim_{h\rightarrow0}\frac{2xh+h^2}{h}$

now we can cancel our h's in the numerator and denominator to get

$\lim_{h\rightarrow 0}2x+h$

then we can just plug 0 in for h and we get our answer

Finding limits of rational functions.

Let

$f(x)=\frac{x^{2}-9}{x+3}$ .

Find

$\lim_{x\rightarrow -3}f(x)$ .

$\infty$

Undefined

Explanation

First factor the numerator to simplify the function.

$x^{2}-9=(x+3)(x-3)$ ,

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

Now

$\lim_{x\rightarrow -3}f(x)$ $=\lim_{x\rightarrow -3}(x-3)$ .

There is no denominator now, and hence no discontinuity. The limit can be found by simply plugging in for .

$=\lim_{x\rightarrow -3}(x-3)=-3-3=-6$ .

Finding limits of rational functions.

Let

$f(x)=\frac{x^{2}-9}{x+3}$ .

Find

$\lim_{x\rightarrow -3}f(x)$ .

$\infty$

Undefined

Explanation

First factor the numerator to simplify the function.

$x^{2}-9=(x+3)(x-3)$ ,

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

Now

$\lim_{x\rightarrow -3}f(x)$ $=\lim_{x\rightarrow -3}(x-3)$ .

There is no denominator now, and hence no discontinuity. The limit can be found by simply plugging in for .

$=\lim_{x\rightarrow -3}(x-3)=-3-3=-6$ .

Limits

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