# L'hospitals rulee major help

• angel_eyez
In summary, Dick is brilliant and has figured out how to solve the problem with just a little bit of knowledge of the Taylor series expansion for e^x.

#### angel_eyez

l'hospitals rulee! major help

1. i don't know how to do this question

2.the right answer is 1/e but I am not getting that, i tried many diff ways, i can't post my work it here because its too long and confusing

## The Attempt at a Solution

Hint

Consider the series expansion for e^[1/x], when x is large.

umm so when x is large 1/x = 0 so that means e^o = 1, I am still lossttt

Do you know the Taylor series for e^y, when y is small?

noo..havent learned it =S

i took the ln of both sides, and then found the derivative, and i got the answer as zero.. so i got ln y= 0
den i used the fact that e^lny = y...there fore y= e^0 =S..

Hmmm, I'm not sure how to do it without recourse to a Taylor series expansion. It's easy to prove though that

e^x=1+x+x^2/2!+x^3/3+...

Have you done Taylor series at all?

noo i haven't :|

kk thnnxx for your help newayzz =)

You suggested that L'hopital's rule should be used in the derivation. I played around with that and couldn't see it, but maybe someone else can work it out.

Let me know when you get the answer.

You should have gotten something like x*ln(x/(x+1)) when you took the log. This is an infinity*0 form. To use L'Hopital you want to write it as 0/0 (or infinity/infinity). So write it as ln(x/(x+1))/(1/x). Now take derivative of the numerator and denominator and send x->infinity.

There's no need for l'Hospital's rule, as this limit is elementary

$$\lim_{x\rightarrow \infty}\left(1-\frac{1}{x+1}\right)^{(x+1)\frac{x}{x+1}} =e^{-\lim_{x\rightarrow \infty}\frac{x}{x+1}}=e^{-1}$$

Dick, Brilliant!

dextercioby said:
There's no need for l'Hospital's rule, as this limit is elementary

$$\lim_{x\rightarrow \infty}\left(1-\frac{1}{x+1}\right)^{(x+1)\frac{x}{x+1}} =e^{-\lim_{x\rightarrow \infty}\frac{x}{x+1}}=e^{-1}$$

That's supposing that you know the Taylor series expansion of e^x. It's nice that you can prove it a completely different way.

christianjb said:
Dick, Brilliant!

Thanks. Wish I could say I invented it, but it's a pretty routine thing to do.

christianjb said:
That's supposing that you know the Taylor series expansion of e^x. It's nice that you can prove it a completely different way.

Pardon me ? I just the definition of "e". What's more elementary than that ??

## 1. What is L'Hospital's rule and when is it used?

L'Hospital's rule is a mathematical technique used to evaluate an indeterminate form of a limit, where both the numerator and denominator approach either 0 or infinity. It is typically used in calculus when other methods, such as direct substitution, are not applicable.

## 2. How does L'Hospital's rule work?

L'Hospital's rule involves taking the derivative of both the numerator and denominator of the function in question, and then evaluating the limit using the new derivatives. This process can be repeated until a valid result is obtained.

## 3. What is an indeterminate form of a limit?

An indeterminate form of a limit is a mathematical expression where the value cannot be determined by simply plugging in the given value. It can take various forms such as 0/0, ∞/∞, or 1^∞.

## 4. Can L'Hospital's rule be used for all indeterminate forms?

No, L'Hospital's rule is only applicable to certain indeterminate forms, such as 0/0 and ∞/∞. It cannot be used for forms like 0*∞ or ∞-∞.

## 5. Are there any limitations to using L'Hospital's rule?

Yes, there are some limitations to using L'Hospital's rule. It can only be used for functions that are differentiable in the given interval, and it may not always provide a valid result. It is important to also consider alternative methods and verify the answer obtained using L'Hospital's rule.