Solve the limit without using L'Hopitals $\lim _{x\to 1}\left(\frac{\sqrt[3]{x+7}-2}{2x^2+3x-5\:}\right)$ $$\lim _{x\to 1}\left(\frac{\sqrt[3]{x+7}-2}{2x^2+3x-5\:}\right)$$
I think I need to do in the numerator by multiplying the difference of cubes.
$\lim _{x\to 1}\left(\frac{\left(\sqrt[3]{x+7}-2\right)\cdot \left(\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4\right)}{\left(2x^2+3x-5\right)\:\cdot \left(\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4\right)}\right)=\lim _{x\to 1}\left(\frac{x+7-8}{\left(2x^2+3x-5\right)\:\cdot \left(\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4\right)}\right)=\lim _{x\to 1}\left(\frac{x-1}{\left(2x^2+3x-5\right)\:\cdot \left(\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4\right)}\right)$
Here's what to do with the denominator?
and whether I chose the idea of a solution?

Hint: Just note that $$2x^2+3x-5=(x-1)(2x+5)$$
