Question

$\lim _{x \rightarrow 0} \lim _{x \rightarrow 0} \frac{2(\tan x-\sin x)-x^{3}}{x^{5}}$ is equal to (A) \(1 / 4\) (B) \(1 / 2\) (C) \(1 / 3\) (D) None of these

Short answer

\(\lim _{x \rightarrow 0} \frac{2(\tan x-\sin x)-x^{3}}{x^{5}}\) (A) \(1 / 4\) (B) \(1 / 2\) (C) \(1 / 3\) (D) None of these Answer: **(D) None of these**

Step-by-step solution

Identify the indeterminate form of the given function

First, we need to identify the indeterminate form of the given function as x approaches 0. Given function: \(\frac{2(\tan x-\sin x)-x^{3}}{x^{5}}\) To find the indeterminate form, replace x with 0 and check the form. Indeterminate form: \(\frac{2(\tan 0-\sin 0)-0^{3}}{0^{5}} = \frac{0}{0}\)

Use x and sin(x) small angle approximations

For small values of x, we know that: 1. \(x \approx \sin{x}\) 2. \(x \approx \tan{x}\) So, we can rewrite the given function as: \(\frac{2(\tan x-\sin x)-x^{3}}{x^{5}} \approx \frac{2(x-\sin x)-x^{3}}{x^{5}}\)

Simplify the given function

Now, simplify the expression: \(\frac{2(x-\sin x)-x^{3}}{x^{5}} = \frac{2x-2\sin x-x^{3}}{x^{5}}\)

Factor out a common x term

Factor out a common x term from both the numerator and the denominator of the function: \(\frac{2x-2\sin x-x^{3}}{x^{5}} = \frac{x(2-2\sin\frac{1}{x}-x^{2})}{x^{5}}\) Now, cancel out an x term from both the numerator and the denominator: \(\frac{x(2-2\sin\frac{1}{x}-x^{2})}{x^{5}} = \frac{2-2\sin\frac{1}{x}-x^{2}}{x^{4}}\)

Calculate the limit

Now, calculate the limit as x approaches 0: \(\lim _{x \rightarrow 0} \frac{2-2\sin\frac{1}{x}-x^{2}}{x^{4}} = \lim _{x \rightarrow 0} \frac{2(1-\sin\frac{1}{x})-x^{2}}{x^{4}}\) Since 1 - sin(x) is very close to 0 as x approaches 0, and \(x^{2}\) vanishes as x approaches 0: \(\lim _{x \rightarrow 0} \frac{2(1-\sin\frac{1}{x})-x^{2}}{x^{4}} \approx \frac{0}{0}\) Thus, we need to apply L'Hopital's rule, which states that if the limit is of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then: \(\lim _{x \rightarrow 0} \frac{f(x)}{g(x)} = \lim _{x \rightarrow 0} \frac{f'(x)}{g'(x)}\) Let: \(f(x) = 2(1-\sin\frac{1}{x}) - x^{2}\) \(g(x) = x^{4}\) Compute the derivatives: \(f'(x) = -2(\frac{-\cos\frac{1}{x}}{x^{2}})-2x\) \(g'(x) = 4x^{3}\) Now, compute the limit of the quotient of the derivatives as x approaches 0: \(\lim _{x \rightarrow 0} \frac{-2(\frac{-\cos\frac{1}{x}}{x^{2}})-2x}{4x^{3}} = \lim _{x \rightarrow 0} \frac{-\cos\frac{1}{x}}{2x}-\frac{1}{2}\) Since cos(u) approaches 1 as u approaches 0, the first term of the limit vanishes as x approaches 0: \(\lim _{x \rightarrow 0} \frac{-\cos\frac{1}{x}}{2x}-\frac{1}{2} = 0-\frac{1}{2} = -\frac{1}{2}\) Thus, the limit of the given function as x approaches 0 is: (A) \(1 / 4\) (B) \(1 / 2\) (C) \(1 / 3\) **(D) None of these**