L'Hopital's Questions

**Paymemoney** · May 2nd 2010, 11:22 PM

Hi

I need some help on the following questions:
1) Use L'Hopital's rule to obtain: $\lim_{x \to 0} \frac{e^{x}-x-1}{3x^2}$

$f'(x) = \frac{3x^{2}*(e^{x}-1)-(e^{x}-x-1)*6x^2}{9x^4}$

$f'(x) = \frac{3x^{2}e^{x} - 6xe^{x}+3x^2+6x}{9x^4}$

$f'(x) = \frac{3x(xe^{x} - 2e^{x} +3x+2}{9x^4}$
$<br /> f'(x) = \frac{(xe^{x} - 2e^{x} +3x+2}{x^3}$

so if x=0, i get 0 however book's answer says $\frac{1}{6}$
2) Use L'Hopital's rule to obtain: $\lim_{x \to \infty} \frac{x+2}{x^2+2x+1}$

$f'(x) = \frac{1}{2x+2} = \frac{1}{2\infty+2}$

The only problem i am having is what does $\infty$ equal to?

P.S

**HallsofIvy** · May 3rd 2010, 12:18 AM

Originally Posted by Paymemoney

Hi

I need some help on the following questions:
1) Use L'Hopital's rule to obtain: $\lim_{x \to 0} \frac{e^{x}-x-1}{3x^2}$

$f'(x) = \frac{3x^{2}*(e^{x}-1)-(e^{x}-x-1)*6x^2}{9x^4}$

$f'(x) = \frac{3x^{2}e^{x} - 6xe^{x}+3x^2+6x}{9x^4}$

$f'(x) = \frac{3x(xe^{x} - 2e^{x} +3x+2}{9x^4}$
$<br /> f'(x) = \frac{(xe^{x} - 2e^{x} +3x+2}{x^3}$

so if x=0, i get 0 however book's answer says $\frac{1}{6}$

This is NOT L'Hopital's rule! You have taken the limit of the derivative of f which is NOT, in general, the limit of f.
(And even for that you would NOT get "0"- you would get $\frac{1}{0}$ which does not exist. Your f' does not have a limit.)

L'Hopital's rule says that if, at x= a, $\frac{f(x)}{g(x)}$ is the indeterminant form " $\frac{0}{0}$ " then $\lim_{x\to a}\frac{f(x)}{g(x)}= \frac{\lim_{x\to a} f'(x)}{\lim_{x\to a} g'(x)}$ provided the limits on the right exist.

That is, you differentiate the numerator and denominator separately NOT using the "quotient law".
Here, the numerator is $e^x- x- 1$ and the denominator is $3x^2$ which are 0 when x= 0, giving the form $\frac{0}{0}$ so we can apply L'Hopital's rule.

The derivative of $e^x- x- 1$ is $e^x- 1$ and the derivative of $3x^2$ is $6x$ . Those are both 0 at x= 0 so apply L'Hopital's rule again.

The derivative of $e^x- 1$ is $e^x$ and the derivative of $6x$ is $6$ which are 1 and 6 at x= 0.

The limit is $\frac{1}{6}$ .

2) Use L'Hopital's rule to obtain: $\lim_{x \to \infty} \frac{x+2}{x^2+2x+1}$

$f'(x) = \frac{1}{2x+2} = \frac{1}{2\infty+2}$

Okay, here you did differentiate numerator and denominator separately- although it should not be labeled "f' ".

The only problem i am having is what does $\infty$ equal to?

P.S

It doesn't equal anything- $\infty$ is not a number, just shorthand for "gets larger without bound."

Here you have a fixed number, 1, over a denominator that gets larger and larger without bound- think of $\frac{1}{100}$ , $\frac{1}{10000}$ , $\frac{1}{10000000}$ , etc. What is happening to those numbers as the denominator gets larger and larger?

**Paymemoney** · May 3rd 2010, 02:13 AM

so for question 2 the book's answer says it 0 how did the get that??

**HallsofIvy** · May 3rd 2010, 04:39 AM

Did you look at the numbers I suggested?

$\frac{1}{100}$ , $\frac{1}{100000}$ , $\frac{1}{100000000000}$ are equal to what in decimal form?

If the numerator of a fraction is a constant, and the denominator gets larger and larger, what happens to the fraction?

**Paymemoney** · May 3rd 2010, 10:28 PM

closer and closer to 0.

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