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

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.meta.math.stackexchange.com/questions/5020/… – Shobhit Sep 29 '13 at 10:49
• Can you find the limit of $(x^2+3x+5)^x$? – Did Sep 29 '13 at 10:54
• @Did the limit that you mention is = 5^0 = 1 right? – IndyZa Sep 29 '13 at 10:59
• Right, and then the limit of $x^x$? – Did Sep 29 '13 at 11:11

$$L=\lim_{x\to0}x\ln(x^2+3x+5)-2\lim_{x\to0}x\ln x=-2\lim_{x\to0}x\ln x$$
Put $y=\frac1x$ and use this
• 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
• @IndyZa, $$\ln \left(\frac1x\right)=\ln(x^{-1})=-\ln x$$ – lab bhattacharjee Sep 29 '13 at 11:14
Hint: $$\frac{\ln y}{\sqrt{y}} \to 0, \quad y \to +\infty.$$