Question

find lim x-->infinity (1+1/x)^(x+2)
answer is given as e^(-2) but I dont know how to get there

Answer 1

Directly substituting yields the indeterminate form \(1^\infty\). Rewriting/rearranging, you can have
\[\begin{align*}\ln L&=\ln\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{x+2}\\
&=\lim_{x\to\infty}\ln\left(1+\frac{1}{x}\right)^{x+2}\\
&=\lim_{x\to\infty}(x+2)\ln\left(1+\frac{1}{x}\right)\\
&=\lim_{x\to\infty}\frac{\ln\left(1+\dfrac{1}{x}\right)}{\dfrac{1}{x+2}}=\frac{0}{0}
\end{align*}\]
Apply L'Hopital's rule:
\[\lim_{x\to\infty}\frac{\ln\left(1+\dfrac{1}{x}\right)}{\dfrac{1}{x+2}}=\lim_{x\to\infty}\frac{\dfrac{-\dfrac{1}{x^2}}{1+\dfrac{1}{x}}}{-\dfrac{1}{(x+2)^2}}=\lim_{x\to\infty}\frac{(x+2)^2}{x^2+x}=1=\ln L~~\iff~~L=e\]

Answer 2

i want to know how to do the question without lh rule

Answer 3

Ah, in that case, recall the limit definition of \(e\):
\[e=\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x\]
Suppose we add and subtract a 2 in the exponent on the right side:
\[e=\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{x+2-2}=\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{x+2}\left(1+\frac{1}{x}\right)^{-2}\]
A property of limits: the limit of a product is the product of the limits, that is,
\[e=\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{x+2}\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{-2}=\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{x+2}\cdot1=e\]

Answer 4

that was awesome man thanks a lot

Answer 5

I don't get why you're told the answer is \(e^{-2}\). The only way you could have that is if the exponent was \(-2x\) in place of \(x\), or if the fraction was \(-\dfrac{2}{x}\).

Answer 6

yw

Answer 7

oh ya the question was incomplete it was a longer limit question and this was th only part not getting simplified by me

Answer 8

the orignal qustion was lim x - infinity ( (x-1)/(x+1) ) ^ x+2

Answer 9

Okay I see. Thanks for clearing that up