evaluate $\lim(1+3x)^{(1/2x)}$.

Question

Evaluate $\lim(1+3x)^{(1/2x)}$ as x approches 0

We know that :

$\lim(1+x)^{(1/x)} =e$ "as x approaches 0 "

I know that we can manipulate this equation to get $e$ to some power

and I tried so many times to get it but I didn't.

Curious to know.

Answer 1

We have $$(1+3x)^{1/(2x)} = \left((1+3x)^{1/(3x)}\right)^{3/2} = \left( (1+y)^{1/y}\right)^{3/2}$$ where $y=3x$. Now conclude what you want.

Answer 2

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

Then $$\ln{(y)}=(\frac{1}{2x})\ln{(1+3x)}\\= \frac{\ln{(1+3x)}}{2x}$$ Now \begin{align}\lim_{x\to 0}\ln{(y)}&=\lim_{x \to 0}\frac{\ln{(1+3x})}{2x} \\&=\underbrace{ \bigg[\frac{0}{0}\bigg]}_\text{Indeterminate form}\end{align} Thus we make use of L'Hospital's Rule \begin{align} \lim_{x \to 0}\ln{(y)}\\ &= \lim_{x\to 0} \frac{\frac{3}{1+3x}}{2}\\ &= \frac{1}{2}(\lim_{x \to 0}\frac{3}{1+3x}) \\&=\frac{3}{2}\end{align}

\begin{align}\lim_{x\to 0}y &= \lim_{x\to 0}e^{\ln{(y)}} \\ &= e^{\lim_{x \to 0}\ln{(y)}} \\ &= e^{\frac{3}{2}}\end{align}

Answer 3

Here is another approach that is a bit more involved

$$ (1+3x)^{\frac{1}{2x}} = e^{\ln(1+3x)/2x} \therefore $$ As x approaches 0, you must now find a value for

$$ \lim_{x \rightarrow 0}{\frac{\ln(1+3x)}{2x}} = ? $$ knowing the definition:

$$ ln(1+x) = \sum_{k=1}^{\infty}\frac{x^k (-1)^{k-1}}{k} $$

$$ \frac{3}{2}+\sum_{k=2}^{\infty}\frac{(3x)^k(-1)^{k-1}}{2kx} $$

Since $x \rightarrow 0$ all other terms will be zero, therefore

$ \lim_{x \rightarrow 0}(1+3x)^{\frac{1}{2x}} = e^{\frac{3}{2}} $