How to calculate defined limit using l'Hospital's rule? I want to calculate the following limit: 

$$\lim_{x\to 0} (1+2x)^{2/(3x+1)} $$ 

It is clear to see that by plugging the value $x=0$ the result is 1. I want to calculate the limit using L'Hospital's rule, by applying $a^x=e^{x\ln(a)}.$ I got : 

$$e^{2\ln(1+2x)/(3x+1)}$$

Then I tried to calculate the power limit by using l'Hospital's rule, and eventually calculate e^(power limit answer). still could not find the correct answer that way. Can somebody help me? Thanks in advance.
L'Hospital's rule works for limits of the form : 

$$\frac{0}{0}$$

$$\frac{\infty}{\infty}$$

etc, which are called undetermined forms.

You cannot apply it on other cases, the limit of the derivatives of the denominator and numerator will not be the same as the limit of the initial expression. 

As you've mentioned, this is a straight-forward limit to calculate, so : 

$$\lim_{x\to 0} (1+2x)^{2/(3x+1)} = 1$$
let $y$ = $(1+2x)^{2/(3x+1)}$ 

$ln$ $y$ = $\frac{2}{(3x+1)}$ $ln$ $(1+2x)$ 

taking the limit then take the Exponential $e$ of the answer 

is that what you do ?! 

because  its not $\frac{0}{0}$ or $\frac{\infty}{\infty}$ the answer will be vary So , 

$\lim_{x\to0}$ $\frac{4}{3(2x+1)}$ apply L'H again to get $\lim_{x\to0}$ $\frac{0}{6}$ = 0 

$e^{0}$ = 1