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GRE Subject Test: Math : L'Hospital's Rule

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Example Question #1681 : Calculus Ii

Evaluate the following limit:

$\lim_{x\rightarrow 2}\frac{x^2-4}{\sin(2-x)}$

Possible Answers:

$\infty$

Correct answer:

Explanation:

When we evaluate the limit using normal methods, we get the indeterminate form $\frac{0}{0}$ . When this happens, we must use L'Hopital's Rule, which states that $\lim_{x\rightarrow n}\frac{f(x)}{g(x)}=\lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}$

Now, we must find the derivatives of the numerator and denominator:

$f'(x)=2x, g'(x)=-\cos(x-4)$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)$

Next, rewrite the limit and evaluate it:

$\lim_{x\rightarrow 2}\frac{2x}{-\cos(2-x)}=-4$

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Example Question #1682 : Calculus Ii

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

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

Possible Answers:

$\infty$

Correct answer:

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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GRE Subject Test: Math : L'Hospital's Rule

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

All GRE Subject Test: Math Resources

Example Questions

Example Question #1681 : Calculus Ii

Example Question #1682 : Calculus Ii

All GRE Subject Test: Math Resources

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