Question

Find $\underset{x\to 0}{lim}\frac{{\sin }^2\left(nx\right)+x^2}{x^2}$

• A. $0$
• B. $n^2+1$
• C. $n+1$
• D. $\frac{1}{n^2+1}$

Answer 1

The correct option is B $n^2+1$

Weneed to find value of ${lim}_{x\to 0}\frac{{\sin }^2\left(nx\right)+x^2}{x^2}$

Divided by highest denominator (divide by $x^2$)

Limit is equal to $={lim}_{x\to 0}[\frac{{\sin }^2\left(nx\right)}{x^2}+\frac{x^2}{x^2}]$

$={lim}_{x\to 0}\frac{{\sin }^2\left(nx\right)}{x^2}+1$

$\frac{{\sin }^2\left(nx\right)}{x^2}={\left(\frac{\sin \left(nx\right)}{x}\right)}^2$

$={lim}_{x\to 0}{\left(\frac{\sin \left(nx\right)}{x}\right)}^2+1$

Apply L' Hospital Rule

$={lim}_{x\to 0}{\left(\frac{\cos \left(nx\right)n}{1}\right)}^2+1$

$={lim}_{x\to 0}{\left(n\cos \left(xn\right)\right)}^2+1$

Put the value $x=0$

$={\left(n\cos (0\cdot n)\right)}^2+1$

$=n^2+1$