If the function $f\left(x\right)$  defined by $f\left(x\right)$ $=\left\{\begin{array}{l}\begin{array}{c}{[1+px]}^{\frac{1}{x}}:x<0\\ q: x=0\\ \frac{{(x+r)}^{1/3}–1}{\sqrt{x+1}–1}:x>0\end{array}\\ \end{array}\right.$  is continuous at $x=0$ , then $p,q,r$  are given by :

$f\left(x\right)$ $=\left\{\begin{array}{l}\begin{array}{c}{[1+px]}^{\frac{1}{x}}:x<0\\ q:x=0\\ \frac{{(x+r)}^{1/3}–1}{\sqrt{x+1}–1}:x>0\end{array}\\ \end{array}\right.$ 

$f\left(0^{–}\right)=$ $\underset{h\to 0}{\lim }$ ${(1-ph)}^{(-1/h)}=$ $\underset{h\to 0}{\lim }$

$f\left(0\right)=q$

$f\left(0^{+}\right)=$ $\underset{h\to 0}{\lim }$ $\frac{{(h+r)}^{1/3}-1}{\sqrt{h+1}-1}$

This limit exists only when $r=1$

Then $f\left(0^{+}\right)=$ $\underset{h\to 0}{\lim }$ $\frac{\left\{\sqrt{h+1}+1\right\}\left[\right(h+1)-1]}{h\left\{{(h+1)}^{2/3}+{(h+1)}^{1/3}+1\right\}}=\frac{2}{3}$

$\Rightarrow r=1,e^p=q=2/3$

$\Rightarrow p=\ln (2/3),q=2/3,r=1$ .