$\lim_{x\to 0} (1+\frac3x+\frac5{x^2})^x$ without L'Hospital

Question

This is what I get for not

$$L = \lim _{x\to0} \left(1+\dfrac3x+\dfrac5{x^2}\right)^x$$ $$\ln(L) = \lim _{x\to 0} \left(x \ln \left( 1+\frac3x+\dfrac5{x^2} \right)\right)$$

I don't know how to prove the $\ln(1+3/x+5/x^2)$ though

Take a look here to write in LATEX.http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference — Shobhit, Sep 29 '13 at 10:49

score 2 · Accepted Answer · edited Apr 13 '17 at 12:21

2

HINT:

$$L=\lim_{x\to0}x\ln(x^2+3x+5)-2\lim_{x\to0}x\ln x=-2\lim_{x\to0}x\ln x$$

Put $y=\frac1x$ and use this

edited Apr 13 '17 at 12:21

Community

answered Sep 29 '13 at 10:54

lab bhattacharjee

I get $x=-2\lim_{x\to\infinity}\frac{ln 1/x}{x} = -2(-1)\lim_{x\to\infinity}\frac{lnx}{x} right?$ – IndyZa Sep 29 '13 at 11:13
1

@IndyZa, $$\ln \left(\frac1x\right)=\ln(x^{-1})=-\ln x$$ – lab bhattacharjee Sep 29 '13 at 11:14

njguliyev · Answer 2 · 2013-09-29T11:03:25.287

0

Hint: $$\frac{\ln y}{\sqrt{y}} \to 0, \quad y \to +\infty.$$

edited Sep 29 '13 at 11:03

answered Sep 29 '13 at 10:50

njguliyev

2 Answers2