Question

Find the limits of $\sqrt[x]{x\frac{a+x}{a-x}},$ when $x=0$.

Answer 1

To find the limits of $\sqrt[x]{x\frac{a+x}{a-x}}$ when $x=0$

i.e. $L=\underset{x\to 0}{lim}\sqrt[x]{x\frac{a+x}{a-x}}$

$=\underset{x\to 0}{lim}{\left\{\frac{a+x}{a-x}\right\}}^{1/x}$ which is in the form of $1^{\infty }$

So, we have to apply another method to find the limit

If the function is of the form $\underset{x\to a}{lim}\left[f\right(x){]}^{g\left(x\right)}$, it turns out that

$\underset{x\to a}{lim}f\left(x\right)=1$ and $\underset{x\to a}{lim}g\left(x\right)=\infty$

Then, this can be transformed into

$=e^{\underset{x\to a}{lim}\left[\right(f\left(x\right)-1)\times g(x\left)\right]}$

Here $f\left(x\right)=\frac{a+x}{a-x}$ and $g\left(x\right)=\frac{1}{x}$

Then,

$L=e^{\underset{x\to 0}{lim}\{(\frac{a+x}{a-x}-1)\times \frac{1}{x}\}}$

$=e^{\underset{x\to 0}{lim}\{\left(\frac{2x}{a-x}\right)\times \frac{1}{x}\}}$

$=e^{\underset{x\to 0}{lim}\left\{\frac{2}{a-x}\right\}}$

$=e^{\frac{2}{a}}$