Question

Solve the limit given on the picture

Answer 1

Pull out the highest power of $n$ from each radical expression, then divide through the numerator and denominator by the largest power of $n$ overall.

The largest power in the cube root is $n^2$; the largest power in the fourth root is $n^{12}$:

$\displaystyle\lim_{n\to\infty}\frac{(n^2-1)^{1/3}+7n^3}{(n^{12}+n+1)^{1/4}-n}=\lim_{n\to\infty}\frac{n^{2/3}\left(1-\dfrac1{n^2}\right)^{1/3}+7n^3}{n^3\left(1+\dfrac1{n^{11}}+\dfrac1{n^{12}}\right)^{1/4}-n}$

Now the largest power in the numerator and denominator is $n^3$, so we get

$=\displaystyle\lim_{n\to\infty}\frac{\dfrac1{n^{7/3}}\left(1-\dfrac1{n^2}\right)^{1/3}+7}{\left(1+\dfrac1{n^{11}}+\dfrac1{n^{12}}\right)^{1/4}+\dfrac1{n^2}}$

Every term containing $n$ approaches 0 as $n\to\infty$, which leaves us with

$=\displaystyle\lim_{n\to\infty}\frac{0(1-0)^{1/3}+7}{(1+0+0)^{1/4}+0}=\frac71=7$