Question

If $\displaystyle \lim_{x \to 0}\dfrac{{{x}^{n}}{{\sin }^{n}}x}{{{x}^{n}}-{{\sin }^{n}}x}$ is non-zero finite, then n is equal to
A. 1
B. 2
C. 3
D. None of these.

Answer 1

In this problem, we have to find the value of n if the given limit is non-zero finite. We know that we cannot directly apply the $x \to 0$ in the limit as the value becomes indeterminate. So, we can use the L ’Hospital Rule, that differentiates numerator and denominator separately, until indeterminate forms exist and then we can substitute $x \to 0$, to get the answer. We can analyse that, for which value of n among the given options is correct for the limit to become non-zero finite.

Complete step-by-step solution:
Here we are given a non-zero finite limit,
$\displaystyle \lim_{x \to 0}\dfrac{{{x}^{n}}{{\sin }^{n}}x}{{{x}^{n}}-{{\sin }^{n}}x}$
Now we can apply $x \to 0$in the above limit, we get
$\Rightarrow \displaystyle \lim_{x \to 0}\dfrac{{{x}^{n}}{{\sin }^{n}}x}{{{x}^{n}}-{{\sin }^{n}}x}=\dfrac{0}{0}$.
We know that the above step is in indeterminate form.
We know, L ‘Hospital Rule states that, when the limit of $\dfrac{f\left( x \right)}{g\left( x \right)}$ is indeterminant, under a certain condition it can be obtained by evaluating the limit of quotient of the derivatives of f and g, i.e., $\dfrac{f'\left( x \right)}{g'\left( x \right)}$. If this result is indeterminate, the procedure can be repeated.
Before that, we can take the given options and substitute the value for n to make the given limit as non-zero finite value.
We can now take option A. as n = 1 and substitute and check for the value as non-zero finite.
$\Rightarrow \displaystyle \lim_{x \to 0}\dfrac{x\sin x}{x-\sin x}=\dfrac{0}{0}$
Now we can apply the L’ Hospital Rule and differentiate the numerator and denominator separately for the given limit.

& \Rightarrow \lim \dfrac{x\sin x}{x-\sin x} \\\ & \Rightarrow \lim \dfrac{x\left( \cos x \right)+\left( 1 \right)\sin x}{1-\cos x} \\\ \end{aligned}$$ We can now apply the limit, we get $$\Rightarrow \displaystyle \lim_{x \to 0}\dfrac{x\left( \cos x \right)+\left( 1 \right)\sin x}{1-\cos x}=\dfrac{0+0}{0}$$ Here we get indeterminate form again, so we can apply the L’ Hospital Rule and differentiate the numerator and denominator again in the above step, we get $$\begin{aligned} & \Rightarrow \lim \dfrac{x\left( \cos x \right)+\left( 1 \right)\sin x}{1-\cos x} \\\ & \Rightarrow \lim \dfrac{x\left( -\sin x \right)+\cos x+\sin x}{0+\sin x} \\\ \end{aligned}$$ We can now apply the limit, we get $$\Rightarrow \displaystyle \lim_{x \to 0}\dfrac{x\left( -\sin x \right)+\cos x+\sin x}{0+\sin x}=\dfrac{0+1+0}{0}$$ Again, we get an indeterminate, so again we can apply the L’ Hospital Rule and differentiate the numerator and denominator, we get $$\begin{aligned} & \Rightarrow \lim \dfrac{x\left( -\sin x \right)+\cos x+\sin x}{0+\sin x} \\\ & \Rightarrow \lim \dfrac{x\left( -\cos x \right)+\left( -\sin x \right)+\cos x+\sin x}{\cos x} \\\ \end{aligned}$$ We can now apply the limit, we get $$\Rightarrow \displaystyle \lim_{x \to 0}\dfrac{x\left( -\cos x \right)+\left( -\sin x \right)+\cos x+\sin x}{\cos x}=\dfrac{0\left( -\cos 0 \right)+\left( \cos 0 \right)+}{\cos 0}=\dfrac{0+1}{1}=1$$ Therefore, we get a non-zero finite as the answer for the given limit, if n = 1. **Therefore, the answer is option A. 1.** **Note:** We should always remember some of the trigonometric degree values, such as $$\sin {{0}^{\circ }}=0,\cos {{0}^{\circ }}=1$$. We should also remember that L ‘Hospital Rule states that, when the limit of $$\dfrac{f\left( x \right)}{g\left( x \right)}$$ is indeterminant, under a certain condition it can be obtained by evaluating the limit of quotient of the derivatives of f and g, i.e., $$\dfrac{f'\left( x \right)}{g'\left( x \right)}$$. If this result is indeterminate, the procedure can be repeated.