Question

Evaluate $\lim_{n \to \infty} \frac{n^{2} }{(n-7)^{2}-6 }$.

Answer 1

$\underline{\text{Solution :}}$

$\displaystyle \mathrm{Now,\: \lim_{n \to \infty} \frac{n^{2}}{(n-7)^{2}-6}}$

$\displaystyle \mathrm{=\lim_{n \to \infty} \frac{n^{2}}{n^{2}-14n+49-6}}$

$\displaystyle \mathrm{=\lim_{n \to \infty} \frac{n^{2 }}{n^{2}-14n+43}}$

$\displaystyle\mathrm{=\lim_{n \to \infty} \frac{\frac{n^{2}}{n^{2}}}{\frac{n^{2}}{n^{2}}-\frac{14n}{n^{2}}+\frac{43}{n^{2}}}}$

$\displaystyle\mathrm{=\lim_{n \to \infty} \frac{1}{1-\frac{14}{n}+\frac{43}{n^{2}}}}$

$\displaystyle\mathrm{=\frac{1}{\mathrm{1-14\displaystyle \mathrm{\lim_{n \to \infty}\frac{1}{n}}+43\displaystyle \mathrm{\lim_{n \to \infty}\frac{1}{n^{2}}}}}}$

$\displaystyle\mathrm{=\frac{1}{1-0+0}}$

$\displaystyle\mathrm{=\frac{1}{1}}$

$\displaystyle\mathrm{=1}$

$\to \boxed{\displaystyle\mathrm{\lim_{n \to \infty} \frac{n^{2}}{(n-7)^{2}-6}=1}}$

$\underline{\text{Rules :}}$

$\displaystyle\mathrm{1.\:\lim_{n \to \infty} \frac{1}{n}=0}$

$\displaystyle\mathrm{2.\:\lim_{n \to \infty} \frac{1}{n^{2}}=0}$

Answer 2

Hey !

Refer the below attachment !

Thanks!