The value of $\underset{x→1}{\lim}\frac{\sqrt[3]{7+x^2}-\sqrt{3+x^2}}{x-1}$ is:
Answer & explanation
Correct answer: option 1
$\underset{x→1}{\lim}\frac{\sqrt[3]{7+x^2}-\sqrt{3+x^2}}{x-1}$
It is an indeterminate $\frac{0}{0}$ form.
On applying L'Hospital's rule, we get :
$\underset{x→1}{\lim}\frac{\frac{1}{3}(7+x^2)^{-2/3}.2x-\frac{1}{2}(3+x^2)^{-1/2}.2x}{1-0}$
$=\frac{1}{3}(8)^{-2/3}.2-\frac{2}{2}(4)^{-1/2}=\frac{2}{3}(2)^{-2}-\frac{2}{2}^{-1}=\frac{2}{3}[\frac{1}{4}]-\frac{1}{2}=\frac{-1}{3}$