First slide

Introduction to limits

Question

If $\lim_{x \to 0} \frac{x^{n} \sin^{n} x}{x^{n} - \sin^{n} x}$ is non-zero finite, then r is equal to

Moderate

A

1
B

2
C

3
D

none of these

Solution

$\begin{array}{l} \lim_{x \to 0} \frac{x^{n} \sin^{n} x}{x^{n} - \sin^{n} x} = \lim_{x \to 0} \frac{x^{n} {(x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - . .)}^{n}}{x^{n} - {(x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - . . . .)}^{n}} \\ = \lim_{x \to 0} \frac{{(x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - . . . . .)}^{n}}{1 - {(1 - \frac{x^{2}}{3!} + \frac{x^{4}}{5!} - . . . . . . .)}^{n}} \\ = \lim_{x \to 0} \frac{x^{n} {(1 - \frac{x^{2}}{3!} + \frac{x^{4}}{5!} - . . . .)}^{n}}{1 - {(1 - \frac{x^{2}}{3!} + \frac{x^{4}}{5!} - . . . .)}^{n}} \\ For n = 2 . \\ = \lim_{x \to 0} \frac{x^{2} {(1 - \frac{x^{2}}{3!} + \frac{x^{4}}{5!} - . . . .)}^{2}}{1 - {(1 - \frac{x^{2}}{3!} + \frac{x^{4}}{5!} - . . . .)}^{2}} \\ = \lim_{x \to 0} \frac{x^{2} {(1 - \frac{x^{2}}{3!} + \frac{x^{4}}{5!} - . . . .)}^{2}}{(2 - \frac{x^{2}}{3!} + \frac{x^{4}}{5!} - . . . .) (\frac{x^{2}}{3!} - \frac{x^{4}}{5!} + . . . . .)} \\ = \frac{1 {(1 - 0 + . . . .)}^{2}}{(2 - 0 + 0) (\frac{1}{3!} - 0 + . . . .)} \\ = 3 \end{array}$

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