Evaluate $\lim\limits_{x \to 0}(\frac{e^2}{(1+ 4x )^{1/2x}})^{1/3x}$

Question

$$\lim\limits_{x \to 0}\left(\frac{e^2}{(1+ 4x )^{\frac1{2x}}}\right)^{\frac1{3x}}=e^{\frac43}$$

I need help with solving this limit. I don't know how to get to the solution. Thanks.

Answer 1

Proceed as follows Let $\displaystyle y=\lim_{x\rightarrow 0}\bigg(\frac{e^2}{(1+4x)^{1/(2x)}}\bigg)^{\frac{1}{3x}}$. Taking log of both sides we have

\begin{array} \ \ln y &=&\lim_{x\rightarrow 0}\frac{1}{3x}\bigg[\ln(e^2)-\ln(1+4x)^{1/2x}\bigg]\\ &=&\displaystyle\lim_{x\rightarrow 0}\frac{2-\frac{1}{2x}\ln(1+4x)}{3x}\\ &=&\displaystyle\lim_{x\rightarrow 0}\frac{4x-\ln(1+4x)}{6x^2}= \text{form}\quad \frac{0}{0} \quad \text{By l'hopital's rule}\\ &=&\displaystyle\lim_{x\rightarrow 0}\frac{4-\frac{4}{1+4x}}{12x} \\ &=&\displaystyle\lim_{x\rightarrow 0}\frac{4}{3(1+4x)}=\frac{4}{3} \end{array} Therefore $y=e^{4/3}$

Answer 2

You can also avoid using L'Hospital. $$A=\left(\frac{e^2}{(1+ 4x )^\frac{1}{2x}}\right)^\frac{1}{3x}\implies \log (A)=\frac{1}{3x}\left(2-\frac1 {2x}\log (1+4x)\right)$$ Now, using Taylor series around $x=0$ $$\log(1+4x)=4 x-8 x^2+\frac{64 x^3}{3}+O\left(x^4\right)$$ $$2-\frac1 {2x}\log (1+4x)=4 x-\frac{32 x^2}{3}+O\left(x^3\right)$$ $$\log(A)=\frac{4}{3}-\frac{32 x}{9}+O\left(x^2\right)$$ Taylor again $$A=e^{\log(A)}=e^{4/3}-\frac{32}{9} e^{4/3} x+O\left(x^2\right)$$ which shows the limit and also how it is approached.

Answer 3

Hint:

Take $A=\left(\frac{e^2}{(1+ 4x )^\frac{1}{2x}}\right)^\frac{1}{3x}$ then $\log A=\frac{1}{3x}\left(\log e^2-\log (1+4x)^{\frac{1}{2x}}\right)$ which is in ($\frac{0}{0}$ form) Now apply L'Hospital's rule to get the limit.