Please solve the problem 65

Please solve 65 th problem 

To solve the given limit problem, we can use L'Hôpital's rule since it involves a $\frac{0}{0}$ indeterminate form. The problem is:

${\lim }_{x\to 0}\frac{x^n{\sin }^{n}x}{x^n-{\sin }^{n}x}$

Given that the limit is a non-zero finite value, let's solve it step-by-step:

1. Apply L'Hôpital's Rule: ${\lim }_{x\to 0}\frac{f(x)}{g(x)}\text{where}f(x)=x^n{\sin }^{n}x\text{and}g(x)=x^n-{\sin }^{n}x$

First derivatives: $f^{\mathrm{\prime }}(x)=\frac{d}{dx}[x^n{\sin }^{n}x]$ $g^{\mathrm{\prime }}(x)=\frac{d}{dx}[x^n-{\sin }^{n}x]$

2. 
   

   
   

   Derivatives: $f^{\mathrm{\prime }}(x)=n{x}^{n-1}{\sin }^{n}x+x^n\cdot n{\sin }^{n-1}x\cdot \cos x$$g^{\mathrm{\prime }}(x)=n{x}^{n-1}-n{\sin }^{n-1}x\cdot \cos x$

3. 
   

   
   

   Simplify the derivatives: ${\lim }_{x\to 0}\frac{n{x}^{n-1}{\sin }^{n}x+n{x}^n{\sin }^{n-1}x\cos x}{n{x}^{n-1}-n{\sin }^{n-1}x\cos x}$

4. 
   

   
   

   Factor out common terms: $={\lim }_{x\to 0}\frac{n{x}^{n-1}{\sin }^{n}x+n{x}^n{\sin }^{n-1}x\cos x}{n{x}^{n-1}-n{\sin }^{n-1}x\cos x}$

Factor $n$ and $x^{n-1}$ out of the numerator and denominator: 

$={\lim }_{x\to 0}\frac{n{x}^{n-1}({\sin }^{n}x+x{\sin }^{n-1}x\cos x)}{n{x}^{n-1}(1-(\sin x\mathrm{/}x{)}^{n-1}\cos x)}$

Since $\sin x\approx x$ as $x\to 0$: $\sin x\approx x\Rightarrow {\sin }^{n}x\approx x^n\Rightarrow \frac{\sin x}{x}\approx 1$

5. 
   
6. 
   
7. Apply these approximations: $={\lim }_{x\to 0}\frac{n{x}^{n-1}(x^n+x\cdot x^{n-1}\cos x)}{n{x}^{n-1}(1-(x\mathrm{/}x{)}^{n-1}\cos x)}$

$={\lim }_{x\to 0}\frac{n{x}^{n-1}(x^n+x^n\cos x)}{n{x}^{n-1}(1-\cos x)}$

$={\lim }_{x\to 0}\frac{n{x}^{n-1}\cdot 2x^n}{n{x}^{n-1}\cdot 0}$

$={\lim }_{x\to 0}\frac{2x^n}{0}$

For the limit to be finite and non-zero, the $x$-terms must cancel out, suggesting $n=1$.

Thus, the value of $n$ must be $1$.

Therefore, the answer is: $(1)1$​

Hope you got your answer!