Question

If $\underset{x\to 0}{lim}\frac{x^n-{\sin }^{n}x}{x-{\sin }^{n}x}$ is non-zero finite, then $n$ must be equal

• A. 4
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is B 1

$\underset{x\to 0}{lim}\frac{x^n-{\sin }^{n}x}{x-{\sin }^{n}x}$
$\text{For n = 0,we have}\underset{x\to 0}{lim}\frac{1-1}{x-1}=0$
$\text{For n = 1,}\underset{x\to 0}{lim}\frac{x-\sin x}{x-\sin x}=1$
$\text{For n = 2,}\underset{x\to 0}{lim}\frac{x^2-{\sin }^{2}x}{x-{\sin }^{2}x}=\underset{x\to 0}{lim}\frac{1-\frac{{\sin }^{2}x}{x^2}}{\frac{1}{x}-\frac{{\sin }^{2}x}{x^2}}.$
$\text{This does not exist.}$
$Forn=3,\underset{x\to 0}{lim}\frac{x^3-{\sin }^{3}x}{x-{\sin }^{3}x}=\underset{x\to 0}{lim}\frac{1-\frac{{\sin }^{3}x}{x^3}}{\frac{1}{x^2}-\frac{{\sin }^{3}x}{x^3}}$
For $n=3$, also given limit does not exist.
Hence, $n=1$