\hspace{-16}\bold{(Q):(1):}\; \mathbf{\lim_{x\rightarrow 0}\left(\frac{b^{x+1}-a^{x+1}}{b-a}\right)^{\frac{1}{x}}=}$\\\\\\ $\bold{(Q):(2):}\;$If $\mathbf{f(x)=}$\begin{cases} \mathbf{\displaystyle \frac{x^2-1}{x^3-1}}\;\;, & \text{ if } \mathbf{x<1} \\\\ \mathbf{\displaystyle \frac{1}{x}\;\;,} & \text{ if } \mathbf{x>1 } \end{cases}\\\\\\ Then $\mathbf{\lim_{x\rightarrow 1^{+}}f(f(x))=}$\\\\\\ $\mathbf{Ans=\frac{2}{3}}$
2621 ) L= e^{\lim_{x\rightarrow 0 }\frac{b^{x+1}-a^{x+1}}{(b-a)x}}
using L Hospitals theorem ,
L= e^{\lim_{x\rightarrow 0 }\frac{b^{x+1}.ln b-a^{x+1}.lna}{(b-a)}}
\Rightarrow L= e^{\lim_{x\rightarrow 0 }\frac{b.ln b-a.lna}{(b-a)}}
L = (\frac{b^{b}}{a^{a}})^{\frac{1}{b-a}}
2622)
for x→1+ f(x)=1/x , therefore f(x) <1 .
hence f(f(x)) for x→1+ =f(x)2 -1f(x)3-1 = x-x31-x3 .
using l' hospitals theorem ,
limx→1+ f(f(x)) = limx→1+x-x31-x3 = 2/3
