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GRE Subject Test: Math : GRE Subject Test: Math

Study concepts, example questions & explanations for GRE Subject Test: Math

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Example Questions

Example Question #1 : L'hospital's Rule

Evaluate:

$\lim_{x\rightarrow 0^{+} } \frac{x \sin x}{2^{x} - 1}$

Possible Answers:

$\frac{1}{\ln2}$

$\ln 2$

Correct answer:

Explanation:

$\lim_{x\rightarrow 0^{+} } x \sin x= 0 \cdot \sin 0 = 0$

and

$\lim_{x\rightarrow 0^{+} } \left ( 2 ^{x} - 1 \right )= 2^{0} -1 = 0$

Therefore, by L'Hospital's Rule, we can find $\lim_{x\rightarrow 0^{+} } \frac{x \sin x}{2^{x} - 1}$ by taking the derivatives of the expressions in both the numerator and the denominator:

$\frac{\mathrm{d} }{\mathrm{d} x}x \sin x = \frac{\mathrm{d} }{\mathrm{d} x} x \cdot \sin x + \frac{\mathrm{d} }{\mathrm{d} x}\sin x \cdot x$

$= 1 \cdot \sin x + \cos x \cdot x = \sin x + x \cos x$

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( 2^{x} - 1 \right ) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( e^{x \ln 2} - 1 \right )$

$= \frac{\mathrm{d} }{\mathrm{d} x} e^{x \ln 2} - \frac{\mathrm{d} }{\mathrm{d} x} 1$

$= \ln 2\cdot e^{x \ln 2} - 0 = \ln 2\cdot 2^{x }$

$\lim_{x\rightarrow 0^{+} } \frac{x \sin x}{2^{x} - 1} = \lim_{x\rightarrow 0^{+} } \frac{\sin x + x \cos x}{ \ln 2\cdot 2^{x }}$

$= \frac{\sin 0 + 0 \cos 0}{ \ln 2\cdot 2^{0 }} = \frac{ 0 + 0 }{ \ln 2\cdot 1} = 0$

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Example Question #1 : L'hospital's Rule

Solve:

${}\lim_{x \to \infty} x^\frac{5}{x}}$

Possible Answers:

e^5

ln(5)

Correct answer:

Explanation:

Substitution is invalid. In order to solve ${}{ \lim_{x \to \infty} x^\frac{5}{x}}$ , rewrite this as an equation.

${} y= { \lim_{x \to \infty} x^\frac{5}{x}}$

Take the natural log of both sides to bring down the exponent.

${} \lim_{x \to \infty} ln(y)= \lim_{x \to \infty} \frac{5}{x} \cdot ln(x)$

${} \lim_{x \to \infty} ln(y)= 5\lim_{x \to \infty} \frac{ln(x)}{x}$

Since ${} \lim_{x \to \infty} \frac{ln(x)}{x}$ is in indeterminate form, ${}\frac{\infty}{\infty}$ , use the L'Hopital Rule.

L'Hopital Rule is as follows:

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

${} \lim_{x \to \infty} \frac{ln(x)}{x} = \lim_{x \to \infty} \frac{\frac{1}{x}}{1} = 0$

This indicates that the right hand side of the equation is zero.

${} \lim_{x \to \infty} ln(y)= 0$

Use the term to eliminate the natural log.

${} \lim_{x \to \infty} e^l^n^(^y^)= e^0$

$\lim_{x \to \infty} y=e^0=1$

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Example Question #2 : L'hospital's Rule

Evaluate the limit using L'Hopital's Rule.

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

Possible Answers:

Undefined

$\infty$

Correct answer:

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.

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Example Question #5 : Euler's Method And L'hopital's Rule

Evaluate the limit using L'Hopital's Rule.

$\lim_{x \to 0} \frac{2x^2}{ln(x)}$

Possible Answers:

$\infty$

Undefined

Correct answer:

Explanation:

$\lim_{x\rightarrow 0}\frac{4x}{\frac{1}{x}}=\lim_{x\rightarrow 0}4x^2$ .

Since the first set of derivatives eliminates an x term, we can plug in zero for the x term that remains. We do this because the limit approaches zero.

This gives us

$4\cdot 0^2=0$ .

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Example Question #3 : Euler's Method And L'hopital's Rule

Evaluate the limit using L'Hopital's Rule.

$\lim_{x\rightarrow \infty }\frac{5x^3+1}{3x^2}$

Possible Answers:

$\frac{5}{3}$

Undefined

$\frac{15}{6}$

$\infty$

Correct answer:

$\infty$

Explanation:

$\lim_{x\rightarrow \infty }\frac{15x^2}{6x}$ .

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

$\lim_{x\rightarrow \infty }\frac{30x}{6}=5x$ .

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

$5\cdot \infty=\infty$ .

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Example Question #4 : Euler's Method And L'hopital's Rule

Calculate the following limit.

$\lim_{x\rightarrow2} \frac{x^2-4}{2x-4}$

Possible Answers:

Correct answer:

Explanation:

To calculate the limit, often times we can just plug in the limit value into the expression. However, in this case if we were to do that we get $\frac00$ , which is undefined.

What we can do to fix this is use L'Hopital's rule, which says

$\lim_{x\rightarrow 2} \frac{f(x)}{g(x)}=\lim_{x\rightarrow 2} \frac{f'(x)}{g'(x)}$ .

So, L'Hopital's rule allows us to take the derivative of both the top and the bottom and still obtain the same limit.

$\lim_{x-\rightarrow 2} \frac{x^2-4}{2x-4}=\lim_{x\rightarrow 2} \frac{2x}{2}$ .

Plug in x=2 to get an answer of .

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Example Question #1 : L'hospital's Rule

Calculate the following limit.

$\lim_{x\rightarrow \infty} \frac{5x^2+2x+1}{x^4+2}$

Possible Answers:

$\infty$

$-\infty$

$\frac{5}{6}$

Correct answer:

Explanation:

If we plugged in the integration limit to the expression in the problem we would get $\frac{\infty}{\infty}$ , which is undefined. Here we use L'Hopital's rule, which is shown below.

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

This gives us,

$\lim_{x\rightarrow \infty} \frac{5x^2+2x+1}{x^4+2}=\lim_{x\rightarrow \infty} \frac{10x+2}{4x^3}$ .

However, even with this simplified limit, we still get $\frac{\infty}{\infty}$ . So what do we do? We do L'Hopital's again!

$\lim_{x\rightarrow \infty} \frac{10x+2}{4x^3}=\lim_{x\rightarrow \infty} \frac{10}{12x^2}$ .

Now if we plug in infitinity, we get 0.

$\frac{10}{\infty}=0$

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Example Question #1 : L'hospital's Rule

Calculate the following limit.

$\lim_{x->-\infty}\frac{e^{-2x}}{x^2}$

Possible Answers:

$-\infty$

$\infty$

$e^{2}$

Correct answer:

$\infty$

Explanation:

If we plugged in $-\infty$ directly, we would get an indeterminate value of $\frac{\infty}{\infty}$ .

We can use L'Hopital's rule to fix this. We take the derivate of the top and bottom and reevaluate the same limit.

$\lim_{x->-\infty}\frac{e^{-2x}}{x^2}=-\frac{e^{-2x}}{x}$ .

We still can't evaluate the limit of the new expression, so we do it one more time.

$\lim_{x->-\infty}-\frac{e^{-2x}}{x}=\lim_{x->-\infty}2e^{-2x}=\infty$

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Example Question #2 : L'hospital's Rule

Find the

$\small \small \small \lim_{x\rightarrow 0}\frac{e^{x^{2}}-ln(1+x)-1}{cosx-sinx-1}$ .

Possible Answers:

$\small 1$

Does Not Exist

$\small e$

$\small 0$

$\small -1$

Correct answer:

$\small 1$

Explanation:

Subbing in zero into $\small \frac{(e^{x^{2}}-ln(1+x)-1)}{cosx-sinx-1}$ will give you $\small \frac{0}{0}$ , so we can try to use L'hopital's Rule to solve.

First, let's find the derivative of the numerator.

$\small e^{x^{2}}$ is in the form $\small a^{u}$ , which has the derivative $\small a^{u}u'lna$ , so its derivative is $\small 2xe^{x^{2}}$ .

$\small ln(1+x)$ is in the form $\small lnu$ , which has the derivative $\small u'/u$ , so its derivative is $\small 1/(1+x)$ .

The derivative of $\small 1$ is $\small 0$ so the derivative of the numerator is $\small 2xe^{x^{2}}-1/(1+x)$ .

In the denominator, the derivative of $\small cosx$ is $\small -sinx$ , and the derivative of $\small sinx$ is $\small cosx$ . Thus, the entire denominator's derivative is $\small -(sinx+cosx)$ .

Now we take the

$\small \lim_{x\rightarrow 0}\frac{2xe^{x^{2}}-1/(x+1)}{-(sinx+cosx)}$ , which gives us $\small \frac{-1}{-1}=1$ .

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Example Question #86 : Calculus

Evaluate the following limit.

$\lim_{x->0}\frac{sin(x)}{ln(x)}$

Possible Answers:

$2\pi$

$\frac{\pi}{2}$

$\infty$

$-\infty$

Correct answer:

Explanation:

If we plug in 0 into the limit we get $\frac{0}{0}$ , which is indeterminate.

We can use L'Hopital's rule to fix this. We can take the derivative of the top and bottom and reevaluate the limit.

$\lim_{x->0}\frac{sin(x)}{ln(x)}=\lim_{x->0}xcos(x)=0$ .

Now if we plug in 0, we get 0, so that is our final limit.

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GRE Subject Test: Math : GRE Subject Test: Math

Study concepts, example questions & explanations for GRE Subject Test: Math

All GRE Subject Test: Math Resources

Example Questions

Example Question #1 : L'hospital's Rule

Example Question #1 : L'hospital's Rule

Example Question #2 : L'hospital's Rule

Example Question #5 : Euler's Method And L'hopital's Rule

Example Question #3 : Euler's Method And L'hopital's Rule

Example Question #4 : Euler's Method And L'hopital's Rule

Example Question #1 : L'hospital's Rule

Example Question #1 : L'hospital's Rule

Example Question #2 : L'hospital's Rule

Example Question #86 : Calculus

All GRE Subject Test: Math Resources

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