$$\lim_{x \to 0} \frac{xe^x - \log(1+x)}{x^2}$$ is equal to-

Explanation

$$\lim_{x \to 0} \frac{xe^x - \log(1+x)}{x^2} = \frac{0}{0}$$ From the L Hospital rule $$\frac{\frac{d}{dx}[xe^x - \log(1+x)]}{\frac{d}{dx}(x)^2}$$ $$\lim_{x \to 0} \frac{xe^x + e^x - \frac{1}{1+x}}{2x}$$ On applying L Hospital rule again $$\lim_{x \to 0} \frac{xe^x + e^x + e^x + \frac{1}{(1+x^2)}}{2}$$ $$= \frac{0+1+1+1}{2} = \frac{3}{2}$$