Evaluate $\lim_{x \rightarrow \infty}\left( \sqrt[3]{x^3+1} -\sqrt[4]{x^4+1}\right)$ Could you help to evaluate $$\lim_{x \rightarrow \infty}\left( \sqrt[3]{x^3+1} -\sqrt[4]{x^4+1}\right)?$$
I tried all methods I know of to no avail
\begin{align*}
\lim_{x \to \infty}{\left(\sqrt[3]{x^3 + 1} - \sqrt[4]{x^4 + 1}\right)} &amp;= \lim_{x \to \infty}{x\left(\sqrt[3]{1 + \frac{1}{x^3}} - \sqrt[4]{1 + \frac{1}{x^4}}\right)}\\
&amp;= \lim_{x \to \infty}{x\left(1 + \frac{1}{3x^3} + O\left(\frac{1}{x^6}\right) -  1 - \frac{1}{4x^4} + O\left(\frac{1}{x^8}\right)\right)}\\
&amp; =\lim_{x \to \infty}{\frac{1}{3x^2} - \frac{1}{4x^3} + O\left(\frac{1}{x^5} \right)}\\
&amp; = 0.
\end{align*}
That's equivalent to $$\lim_{x \rightarrow 0} \frac{(x^3+1)^{1/3} -(x^4+1)^{1/4}}{x},$$ to which you may apply L'hopital.