Solve the limit given on the picture

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

The largest power in the cube root is [tex]n^2[/tex]; the largest power in the fourth root is [tex]n^{12}[/tex]:

[tex]\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}[/tex]

Now the largest power in the numerator and denominator is [tex]n^3[/tex], so we get

[tex]=\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}}[/tex]

Every term containing [tex]n[/tex] approaches 0 as [tex]n\to\infty[/tex], which leaves us with

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