Question

$\lim_{x \to 0} \frac{x-sinx}{x^3}$

Answer 1

1/6

Answer 2

We are asked to evaluate the limit $\lim_{x \to 0} \frac{x-\sin x}{x^3}$.

As $x \to 0$, the numerator $x - \sin x \to 0 - \sin 0 = 0$.
As $x \to 0$, the denominator $x^3 \to 0^3 = 0$.
The limit is of the indeterminate form $\frac{0}{0}$. We can use either L'Hôpital's Rule or Taylor series expansion.

Method 1: Using L'Hôpital's Rule

Applying L'Hôpital's Rule:

$$\lim_{x \to 0} \frac{x-\sin x}{x^3} = \lim_{x \to 0} \frac{\frac{d}{dx}(x-\sin x)}{\frac{d}{dx}(x^3)} = \lim_{x \to 0} \frac{1-\cos x}{3x^2}$$

This is still of the form $\frac{0}{0}$. Applying L'Hôpital's Rule again:

$$= \lim_{x \to 0} \frac{\frac{d}{dx}(1-\cos x)}{\frac{d}{dx}(3x^2)} = \lim_{x \to 0} \frac{-(-\sin x)}{6x} = \lim_{x \to 0} \frac{\sin x}{6x}$$

This is still of the form $\frac{0}{0}$. Applying L'Hôpital's Rule a third time:

$$= \lim_{x \to 0} \frac{\frac{d}{dx}(\sin x)}{\frac{d}{dx}(6x)} = \lim_{x \to 0} \frac{\cos x}{6}$$

Now, substitute $x=0$:

$$= \frac{\cos 0}{6} = \frac{1}{6}$$

Method 2: Using Taylor Series Expansion

The Taylor series expansion of $\sin x$ around $x=0$ (Maclaurin series) is:

$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots$$

Substitute this into the expression:

$$\frac{x-\sin x}{x^3} = \frac{x - \left(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots \right)}{x^3}$$ $$= \frac{x - x + \frac{x^3}{3!} - \frac{x^5}{5!} + \dots}{x^3}$$ $$= \frac{\frac{x^3}{6} - \frac{x^5}{120} + \dots}{x^3}$$

For $x \neq 0$, we can divide the numerator by $x^3$:

$$= \frac{1}{6} - \frac{x^2}{120} + \frac{x^4}{5040} - \dots$$

Now, take the limit as $x \to 0$:

$$\lim_{x \to 0} \left(\frac{1}{6} - \frac{x^2}{120} + \frac{x^4}{5040} - \dots \right) = \frac{1}{6} - 0 + 0 - \dots = \frac{1}{6}$$

Both methods give the same result.