Question

Find the limit.

Answer 1

\[\lim_{x \rightarrow \infty}(1+\frac{ 1 }{ x })^{x^2}\]

Answer 2

Well you have a few options here.
You can use an exponent rule and limit rule, writing it this way,\[\large\rm \lim_{x \rightarrow \infty}\left(1+\frac{ 1 }{ x }\right)^{x^2}=\lim_{x \rightarrow \infty}\left[\left(1+\frac{ 1 }{ x }\right)^{x}\right]^2=\left[\color{orangered}{\lim_{x \rightarrow \infty}\left(1+\frac{ 1 }{ x }\right)^{x}}\right]^2\]And then apply the definition of \(\large\rm \color{orangered}{e}\) to this orange part,

The other option:
You can use logarithms, and L'Hopital's Rule and all that jazz.

Answer 3

\[lny = \lim_{x \rightarrow \infty}x^2\ln(\frac{ x+1 }{ x })\]
@zepdrix like that?

Answer 4

I would avoid doing common denominator maybe..
I think it'll make your derivatives easier.
But yes.

Answer 5

\[lny=\ln \frac{ x+1 }{ x }/ \frac{ 1 }{x^2 }\]

Answer 6

mhm

Answer 7

lny = \[\lim_{x \rightarrow \infty}\frac{ \frac{ -1 }{ x^2+x } }{ \frac{ -2 }{ x^3 } }\]
How would I simplify this? @zepdrix

Answer 8

Flip the bottom one I guess\[\large\rm \frac{-1}{x^2+x}\cdot \frac{x^3}{-2}\]

Answer 9

\[\frac{ -1 }{ x(x+1) }*\frac{ x^3 }{ -2 }= \frac{ -1 }{ x+1 }*\frac{ x^2 }{ -2 }\]
like that?

Answer 10

\[\ln y = \lim_{x \rightarrow \infty}x\]
\[y = e^\infty = \infty \]
e to the power of infinity is infinity right?

Answer 11

@zepdrix

Answer 12

That's not right :(
Something must've gotten mixed up in your derivative. I didn't check that.

Answer 13

\[\large\rm \frac{d}{dx}\ln\left(1+\frac{1}{x}\right)=\frac{\frac{-1}{x^2}}{1+\frac{1}{x}}\]
\[\large\rm \frac{d}{dx}\frac{1}{x^2}=\frac{-2}{x^3}\]

Answer 14

So then,\[\large\rm \lim_{x\to\infty}\frac{\ln\left(1+\frac1x\right)}{\frac{1}{x^2}}\quad=\quad \lim_{x\to\infty}\frac{\frac{\frac{-1}{x^2}}{1+\frac1x}}{\frac{-2}{x^3}}\quad=\quad \lim_{x\to\infty}\frac{1}{1+\frac1x}\cdot\frac{-1}{x^2}\cdot\frac{x^3}{-2}\]Something like that?
I might still be making a mistake lol

Answer 15

OH OH OH, the original exponent is 2x, not x^2

Answer 16

That's what is throwing everything off,
AHHH now I see.

Answer 17

Can I do it like this without simply solving?

let t = \[\frac{ x }{ 2 }\]
\[y = \lim_{x \rightarrow \infty}e^t = \infty\] @zepdrix

Answer 18

We can use the base graph of \[e^x\]

Answer 19

x / 2 is from getting the derivative without using common denominators

Answer 20

The answer is infinity

Answer 21

No the answer is e^2

Answer 22

Oh I guess I looked at the problem wrong :(
Ugh I thought you posted this:\[\large\rm \lim_{x\to\infty}\left(1+\frac1x\right)^{2x}\]

Answer 23

fsdajfkladsjfgklasgjds;l

Answer 24

lol