GRE Quantitative Reasoning - Limits

Evaluate the limit using L'Hopital's Rule.

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

Undefined

$\infty$

Explanation

L'Hopital's Rule is used to evaluate complicated limits. The rule has you take the derivative of both the numerator and denominator individually to simplify the function. In the given function we take the derivatives the first time and get

$\lim_{x\rightarrow \infty }\frac{2x}{e^x}$ .

This still cannot be evaluated properly, so we will take the derivative of both the top and bottom individually again. This time we get

$\lim_{x\rightarrow \infty }\frac{2}{e^x}$ .

Now we have only one x, so we can evaluate when x is infinity. Plug in infinity for x and we get

$\frac{2}{e^\infty }$ and $e^\infty=\infty$ .

So we can simplify the function by remembering that any number divided by infinity gives you zero.

Evaluate the limit using L'Hopital's Rule.

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

Undefined

$\infty$

Explanation

$\lim_{x\rightarrow \infty }\frac{2x}{e^x}$ .

This still cannot be evaluated properly, so we will take the derivative of both the top and bottom individually again. This time we get

$\lim_{x\rightarrow \infty }\frac{2}{e^x}$ .

Now we have only one x, so we can evaluate when x is infinity. Plug in infinity for x and we get

$\frac{2}{e^\infty }$ and $e^\infty=\infty$ .

So we can simplify the function by remembering that any number divided by infinity gives you zero.

Evaluate the following limit

$\small \lim_{x\to 0} \frac{\cos x-(x-1)^2}{x^3-3x}$

if possible.

$\small -\frac{2}{3}$

$\small -1$

Limit does not exist

$\small 1$

Explanation

If we try to directly plug in the limit value into the function, we get

$\small \small \frac{\cos 0-(0-1)^2}{0^3-3\cdot 0}=\frac{1-1}{0-0}=\frac{0}{0}$

Because the limit is of the form $\small \frac{0}{0}$ , we can apply L'Hopital's rule to "simplify" the limit to

$\small \small \lim_{x\to 0} \frac{\cos x-(x-1)^2}{x^3-3x}=\lim_{x\to 0} \frac{\sin x-2(x-1)}{3x^2-3}$ .

Now if we directly plug in 0 again, we get

$\small \small \small \lim_{x\to 0} \frac{\cos x-(x-1)^2}{x^3-3x}=\lim_{x\to 0} \frac{\sin x-2(x-1)}{3x^2-3}=\frac{0-2(0-1)}{0-3}=-\frac{2}{3}$ .

Find the limit if it exists

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

Hint: Use L'Hospital's rule

$-\frac{1}{2}$

$\frac{1}{2}$

Explanation

Directly evaluating for yields the indeterminate form

$\frac{sin(0)-0}{0^2}=\frac{0-0}{0}=\frac{0}{0}$

we are able to apply L'Hospital's rule which states that if the limit is in indeterminate form when evaluated, then

$\lim_{x\rightarrow a}\frac{f(a)}{g(a)}=\lim_{x\rightarrow a}\frac{f'(a)}{g'(a)}$

As such the limit in the problem becomes

$\lim_{x\rightarrow 0}\frac{sin(x)-x}{x^2}=\lim_{x\rightarrow 0}\frac{\frac{\mathrm{d} }{\mathrm{d} x}[sin(x)-x]}{\frac{\mathrm{d} }{\mathrm{d} x}[x^2]}=\lim_{x\rightarrow 0}\frac{cos(x)-1}{2x}$

Evaluating for again yields the indeterminate form

$\frac{cos(x)-1}{2x}=\frac{cos(0)-1}{2*0}=\frac{1-1}{0}=\frac{0}{0}$

So we apply L'Hospital's rule again

$\lim_{x\rightarrow 0}\frac{cos(x)-1}{2x}=\lim_{x\rightarrow 0}\frac{\frac{\mathrm{d} }{\mathrm{d} x}[cos(x)-1]}{\frac{\mathrm{d} }{\mathrm{d} x}[2x]}=\lim_{x\rightarrow 0}\frac{-sin(x)}{2}$

Evaluating for yields

$\frac{-sin(0)}{2}=\frac{0}{2}=0$

As such

$\lim_{x\rightarrow 0}\frac{-sin(x)}{2}=0$

and thus

$\lim_{x\rightarrow 0}\frac{sin(x)-x}{x^2}=0$

Evaluate the following limit:

$\lim_{x\rightarrow 3} \frac{3-x}{x^2-9}$

$-\frac{1}{6}$

$\frac{1}{6}$

The limit does not exist

$\infty$

Explanation

When we evaluate the limit using normal methods, we arrive at the indeterminate form $\frac{0}{0}$ . When this occurs, to evaluate the limit, we must use L'Hopital's Rule, which states that

$\lim_{x\rightarrow n}\frac{ f(x)}{g(x)}=\lim_{x\rightarrow n} \frac{f'(x)}{g'(x)}$

So, we must find the derivative of the top and bottom functions:

The derivatives were found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Now, rewrite the limit and evaluate it:

$\lim_{x\rightarrow 3}-\frac{1}{2x}=-\frac{1}{6}$

Use l'Hopital's rule to find the limit:

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

$\infty$

Explanation

The first thing we always have to do is to check that l'Hopital's rule is actually applicable when we want to use it.

$\frac{1^2-3\cdot1+2}{1-1} = \frac{0}{0}$

So it is applicable here.

We take the derivative of the top and bottom, and get

$\lim_{x \to 1} \frac{x^2-3x+2}{x-1} = \lim_{x \to 1} \frac{2x-3}{1}$

and now we can safely plug in x=1 and get that the limit equals

$2\cdot 1 -3 = -1$ .

Evaluate the following limit

$\small \lim_{x\to 0} \frac{\cos x-(x-1)^2}{x^3-3x}$

if possible.

$\small -\frac{2}{3}$

$\small -1$

Limit does not exist

$\small 1$

Explanation

If we try to directly plug in the limit value into the function, we get

$\small \small \frac{\cos 0-(0-1)^2}{0^3-3\cdot 0}=\frac{1-1}{0-0}=\frac{0}{0}$

Because the limit is of the form $\small \frac{0}{0}$ , we can apply L'Hopital's rule to "simplify" the limit to

$\small \small \lim_{x\to 0} \frac{\cos x-(x-1)^2}{x^3-3x}=\lim_{x\to 0} \frac{\sin x-2(x-1)}{3x^2-3}$ .

Now if we directly plug in 0 again, we get

$\small \small \small \lim_{x\to 0} \frac{\cos x-(x-1)^2}{x^3-3x}=\lim_{x\to 0} \frac{\sin x-2(x-1)}{3x^2-3}=\frac{0-2(0-1)}{0-3}=-\frac{2}{3}$ .

Evaluate the following limit:

$\lim_{x\rightarrow 3} \frac{3-x}{x^2-9}$

$-\frac{1}{6}$

$\frac{1}{6}$

The limit does not exist

$\infty$

Explanation

When we evaluate the limit using normal methods, we arrive at the indeterminate form $\frac{0}{0}$ . When this occurs, to evaluate the limit, we must use L'Hopital's Rule, which states that

$\lim_{x\rightarrow n}\frac{ f(x)}{g(x)}=\lim_{x\rightarrow n} \frac{f'(x)}{g'(x)}$

So, we must find the derivative of the top and bottom functions:

The derivatives were found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Now, rewrite the limit and evaluate it:

$\lim_{x\rightarrow 3}-\frac{1}{2x}=-\frac{1}{6}$

Find the limit if it exists

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

Hint: Use L'Hospital's rule

$-\frac{1}{2}$

$\frac{1}{2}$

Explanation

Directly evaluating for yields the indeterminate form

$\frac{sin(0)-0}{0^2}=\frac{0-0}{0}=\frac{0}{0}$

we are able to apply L'Hospital's rule which states that if the limit is in indeterminate form when evaluated, then

$\lim_{x\rightarrow a}\frac{f(a)}{g(a)}=\lim_{x\rightarrow a}\frac{f'(a)}{g'(a)}$

As such the limit in the problem becomes

$\lim_{x\rightarrow 0}\frac{sin(x)-x}{x^2}=\lim_{x\rightarrow 0}\frac{\frac{\mathrm{d} }{\mathrm{d} x}[sin(x)-x]}{\frac{\mathrm{d} }{\mathrm{d} x}[x^2]}=\lim_{x\rightarrow 0}\frac{cos(x)-1}{2x}$

Evaluating for again yields the indeterminate form

$\frac{cos(x)-1}{2x}=\frac{cos(0)-1}{2*0}=\frac{1-1}{0}=\frac{0}{0}$

So we apply L'Hospital's rule again

$\lim_{x\rightarrow 0}\frac{cos(x)-1}{2x}=\lim_{x\rightarrow 0}\frac{\frac{\mathrm{d} }{\mathrm{d} x}[cos(x)-1]}{\frac{\mathrm{d} }{\mathrm{d} x}[2x]}=\lim_{x\rightarrow 0}\frac{-sin(x)}{2}$

Evaluating for yields

$\frac{-sin(0)}{2}=\frac{0}{2}=0$

As such

$\lim_{x\rightarrow 0}\frac{-sin(x)}{2}=0$

and thus

$\lim_{x\rightarrow 0}\frac{sin(x)-x}{x^2}=0$

Use l'Hopital's rule to find the limit:

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

$\infty$

Explanation

The first thing we always have to do is to check that l'Hopital's rule is actually applicable when we want to use it.

$\frac{1^2-3\cdot1+2}{1-1} = \frac{0}{0}$

So it is applicable here.

We take the derivative of the top and bottom, and get

$\lim_{x \to 1} \frac{x^2-3x+2}{x-1} = \lim_{x \to 1} \frac{2x-3}{1}$

and now we can safely plug in x=1 and get that the limit equals

$2\cdot 1 -3 = -1$ .

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