Find the Limit of a Function - Pre-Calculus

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Question

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

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

Answer

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.

This can be illustrated by thinking of small negative numbers.

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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Question

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)?

Answer

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 !

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Question

Evaluate the limit below:

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

Answer

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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Question

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

Answer

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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Question

Evaluate the following limit.

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

Answer

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.

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Question

Find the limit as x approaches infinity

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

Answer

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.

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Question

Solve the limit as approaches infinity.

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

Answer

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.

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

$\frac{-3}{13}$

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Question

Find the limit

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

Answer

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.

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Question

Solve the following limit:

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

Answer

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

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Question

Finding limits of rational functions.

Let

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

Find

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

Answer

First factor the numerator to simplify the function.

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

so

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

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 .

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

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Question

The Michaelis-Menten equation is important in chemical kinetics. Suppose we are given the following equation relating K (reaction rate) and C (concentration):

$K(C) = \tfrac{1000C}{500 + C}$

Determine:

$\lim_{C \rightarrow \infty } K(C)$

Answer

There are a number of ways to solve this. Either we can solve by finding what K(C) evaluates to for larger values of C and see where they converge, i.e. :

K(10000) = 952.38

K(1000000) = 999.5

etc...

And we see that K(C) approaches 1000 for larger concentrations, C.

Or we can notice that the dominant term in the numerator is 1000C; dominant term in the denominator is C.

1000C / C = 1000, which will ultimately be our limit.

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Question

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

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

Answer

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

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

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Question

Evaluate

$\lim_{-3}\frac{\frac{1}{x}+\frac{1}{3}}{1+\frac{2}{x+1}}$ .

Answer

Find a common denominator for both the upper and lower expressions and then simplify:

$\ \lim_{x\rightarrow -3}\frac{\frac{1}{x}+\frac{1}{3}}{1+\frac{2}{x+1}}\ \=\lim_{x\rightarrow -3}\frac{\frac{3+x}{3x}}{\frac{x+1+2}{x+1}}\ \=\lim_{x\rightarrow -3}\frac{3+x}{3x}\times\frac{x+1}{3+x}\ \=\lim_{x\rightarrow -3}\frac{x+1}{3x}\ \=\frac{-3+1}{3(-3)}\ \=\frac{-2}{-9}\ \=\frac{2}{9}$

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Question

Find the limit of the function:

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

Answer

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: $-\infty$

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Question

Find the limit.

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

Answer

In the unsimplified form, the limit does not exist; however, the numerator can be factored and simplified.

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

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

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Question

Find the following limit:

$\lim_{x\rightarrow \infty} \frac{3x^2+1}{2x^2}$

Answer

To solve, simply realize you are dealing with a limit whose numerator and demominator have the same max power. Thus the limit is simply the division of their coefficient.

$\frac{3x^2+1}{2x^2}\approx \frac{3x^2}{2x^2}=\frac{3}{2}$

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Question

Find the limit: $\lim_{x \to 6} \frac{x-3}{3}$

Answer

The first and only step is to substitute the value of into the function. Since there is not a zero denominator or an indeterminate form, we do not have to worry about L'Hopital or an undefined limit.

$\lim_{x \to 6} \frac{x-3}{3} = \frac{6-3}{3} = \frac{3}{3} = 1$

The limit will approach to .

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Question

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

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

Answer

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.

This can be illustrated by thinking of small negative numbers.

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.

Compare your answer with the correct one above

Question

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)?

Answer

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 !

Compare your answer with the correct one above

Question

Evaluate the limit below:

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

Answer

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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