Question

Let $f:R\to R$ be such that f(1) =2, f'(1)= 4 then the ${lim}_{x\to 0}[\frac{f(1+x+x^2)+xf(1+x)}{f\left(1\right)}{]}^{\frac{1}{x(x+1)}}$ = _______

• A. $e$
• B. $e^{\frac{1}{2}}$
• C. $e^2$
• D. $e^3$

Answer 1

The correct option is A $e$

Since indetermined form is $1^{\infty }$

So,

${lim}_{x⟶0}{\left[\frac{f(1+x+x^2)+xf(1+x)-1}{f\left(1\right)}\right]}^{\frac{1}{x(x+1)}}={lim}_{x⟶0}{e}^{\frac{1}{x(x+1)}\left[\frac{f(1+x+x^2)+xf(1+x)-1}{f\left(1\right)}\right]}={lim}_{x⟶0}{e}^{\frac{1}{x(x+1)}\frac{xf(1+x)}{f\left(1\right)}}=e^1=e$