Question

Find limit x tends to 0 (x-tan^-1 x) ÷x^3

Answer 1

1/3

Answer 2

Let the given limit be $L$.

$L = \lim_{x \to 0} \frac{x - \tan^{-1} x}{x^3}$

As $x \to 0$, the numerator $x - \tan^{-1} x \to 0 - \tan^{-1}(0) = 0 - 0 = 0$. As $x \to 0$, the denominator $x^3 \to 0^3 = 0$. The limit is of the indeterminate form $\frac{0}{0}$. We can apply L'Hopital's Rule.

Applying L'Hopital's Rule, we differentiate the numerator and the denominator with respect to $x$: Derivative of the numerator: $\frac{d}{dx}(x - \tan^{-1} x) = 1 - \frac{1}{1+x^2}$. Derivative of the denominator: $\frac{d}{dx}(x^3) = 3x^2$.

So, the limit becomes: $L = \lim_{x \to 0} \frac{1 - \frac{1}{1+x^2}}{3x^2}$

Now, simplify the numerator: $1 - \frac{1}{1+x^2} = \frac{(1+x^2) - 1}{1+x^2} = \frac{x^2}{1+x^2}$

Substitute this back into the limit expression: $L = \lim_{x \to 0} \frac{\frac{x^2}{1+x^2}}{3x^2}$ $L = \lim_{x \to 0} \frac{x^2}{(1+x^2)(3x^2)}$

Since $x \to 0$, $x \neq 0$, so we can cancel the $x^2$ term from the numerator and the denominator: $L = \lim_{x \to 0} \frac{1}{3(1+x^2)}$

Now, substitute $x=0$ into the expression: $L = \frac{1}{3(1+0^2)} = \frac{1}{3(1)} = \frac{1}{3}$

Alternatively, using Taylor series expansion: The Taylor series expansion of $\tan^{-1} x$ around $x=0$ is $\tan^{-1} x = x - \frac{x^3}{3} + \frac{x^5}{5} - \dots$ for $|x| \le 1$. Substitute this into the numerator: $x - \tan^{-1} x = x - \left(x - \frac{x^3}{3} + \frac{x^5}{5} - \dots\right) = \frac{x^3}{3} - \frac{x^5}{5} + \dots$

Now divide by $x^3$: $\frac{x - \tan^{-1} x}{x^3} = \frac{\frac{x^3}{3} - \frac{x^5}{5} + \dots}{x^3} = \frac{1}{3} - \frac{x^2}{5} + \dots$

Take the limit as $x \to 0$: $\lim_{x \to 0} \left(\frac{1}{3} - \frac{x^2}{5} + \dots\right) = \frac{1}{3} - 0 + 0 - \dots = \frac{1}{3}$

Both methods give the same result.

The limit is of the form $0/0$. Applying L'Hopital's rule: $\lim_{x \to 0} \frac{x - \tan^{-1} x}{x^3} = \lim_{x \to 0} \frac{\frac{d}{dx}(x - \tan^{-1} x)}{\frac{d}{dx}(x^3)}$ $= \lim_{x \to 0} \frac{1 - \frac{1}{1+x^2}}{3x^2}$ $= \lim_{x \to 0} \frac{\frac{1+x^2-1}{1+x^2}}{3x^2}$ $= \lim_{x \to 0} \frac{x^2}{(1+x^2)(3x^2)}$ Cancel $x^2$ (since $x \neq 0$ in the limit): $= \lim_{x \to 0} \frac{1}{3(1+x^2)}$ Substitute $x=0$: $= \frac{1}{3(1+0^2)} = \frac{1}{3}$