Pre-Calculus - Find the Limit of a Function

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

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

$\frac{13}{2}$

$\infty$

$\frac{3}{2}$

$\frac{9}{2}$

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

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.

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 .

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

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

Evaluate the following limit.

$\lim_{x\rightarrow 0}\frac{x^2+3x}{3x^2+4x}$

$\frac{3}{4}$

$\infty$

$-\infty$

$\frac{1}{3}$

$\textup{Does not exist}$

Explanation

The function has a removable discontinuity at . Once a factor of is "divided out" the resultant function is $\frac{x+3}{3x+4}$ , which evaluates to $\frac{3}{4}$ as approaches 0.

Let $f(x)=\frac{x+7}{x^{2}-2}$ .

Find $\lim_{x\rightarrow 2}$ .

$\frac{9}2{}$

Undefined

$-\frac{3}{4}$

Explanation

To find $\lim_{x\rightarrow 2}$ here, you need only plug in for :

$f(x)=\frac{x+7}{x^{2}-2}=\frac{2+7}{2^{2}-2}=\frac{9}{2}$

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 !

Solve the limit as approaches infinity.

$\lim_{x\rightarrow \infty }\frac{3x^4-2^7x^2+4x}{5x^2-13x^4+5x}$

$\frac{-3}{13}$

$-\infty$

$\infty$

$\frac{2^7}{5}$

Explanation

For our equation the orders are the same in x for the numerator and denominator (both 4). So we divide the coefficients of the highest order x terms to get our limit and we get

$\frac{-3}{13}$

Find the limit of the function:

$\lim_{x\rightarrow 5^{+}} \frac{5}{5-x}$

$-\infty$

$\infty$

$\frac{1}{5}$

The limit can't be determined.

Explanation

Substituting the value of will yield $\frac{5}{0}$ , which is not in indeterminate form. Therefore, L'Hopital's rule cannot be used.

The question asks for the limit as x approaches to five from the right side in the graph. As the graph approaches closer and closer to , the y-value decreases to negative infinity and will never touch .

The correct answer is: $-\infty$

Find the Limit of a Function

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