Question

$\underset{x\to 1}{lim}\frac{\sqrt[3]{37+x^3}-\sqrt{3+x^2}}{x-1}$ is equal to

Answer 1

${lim}_{x\to 1}\frac{{(7+x^3)}^{\frac{1}{3}}-{(3+x^2)}^{\frac{1}{2}}}{x-1}$

Using L=Hopital rule

$\frac{1\times 3x^2}{3{(7+x^3)}^{\frac{2}{3}}}-\frac{1\times 2x}{2{(3+x^2)}^{\frac{1}{2}}}x\to 1\frac{1\times 3}{3\times {(7+1)}^{\frac{2}{3}}}-\frac{2}{2{(3+1)}^{\frac{1}{2}}}\frac{3}{12}-\frac{2}{4}\frac{1}{4}-\frac{1}{2}\frac{-1}{4}$