# Where am I going wrong - limit of sequence

#### Sophdof1

##### New member
Why am I not getting 1/e?? Instead I get e?

I am so stressed as I thought I had finally found the answer. Does anyone see the flaw in my solution? I don't see where I'm going wrong.

Help would be so apreciated.

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#### Jomo

##### Elite Member
When using l'hopital's rule instead of taking the derivative of ln(1/(1+1/x)) you just took the derivative of ln(1+1/x)

• Sophdof1

#### ksdhart2

##### Senior Member

$$\displaystyle \lim_{x \to \infty} \frac{\ln \left( \frac{1}{1+ \frac{1}{x}} \right)}{\frac{1}{x}}$$

But after that you lose me completely. You write that you're applying L'Hopital's Rule, but that doesn't look like any application of the rule I've ever seen. You should end up with:

$$\displaystyle \lim_{x \to \infty} \frac{\frac{d}{dx} \left[ \ln \left( \frac{1}{1+ \frac{1}{x}} \right) \right]}{\frac{d}{dx} \left[ \frac{1}{x} \right]}$$

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

When you simply that down, you'll get:

$$\displaystyle \lim_{x \to \infty} \ln(y(x)) = -1 \implies \lim_{x \to \infty} y(x) = e^{-1} = \frac{1}{e}$$

#### pka

##### Elite Member
Why am I not getting 1/e?? Instead I get e?
I am so stressed as I thought I had finally found the answer. Does anyone see the flaw in my solution? I don't see where I'm going wrong.
Help would be so apreciated.
I am not at all sure what exactly you are asking. If it is to derive the limit $$\displaystyle \mathop {\lim }\limits_{n \to \infty } {\left( {\frac{n}{{n + 1}}} \right)^n} = \frac{1}{e}$$ or to prove it?
Here is a general principal: $$\displaystyle \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{a}{{n + b}}} \right)^{cn}} = {e^{ac}}$$.
If we have proven (or were given) that rule then note that $$\displaystyle \left( {\frac{n}{{n + 1}}} \right) = \left( {1 + \frac{{ - 1}}{{n + 1}}} \right)$$,
hence $$\displaystyle \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{ - 1}}{{n + 1}}} \right)^n} = {e^{ - 1}}$$ (i.e. $$\displaystyle a=-1$$).

If you are asking to derive the principal then that requires an entire section of an advanced text.

#### pka

##### Elite Member
Here is a general principal: $$\displaystyle \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{a}{{n + b}}} \right)^{cn}} = {e^{ac}}$$.
If you are asking to derive the principal then that requires an entire section of an advanced text.
I dug through old computer files (from twenty years ago). Here is a file from a lecture on the general principal.
Because it is converted from slides, you should use the "page down" feature to read the file.

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