Question

The value of $\left(\lim _{x \rightarrow 0}\left[\frac{100 x}{\sin x}\right]+\left[\frac{99 \sin x}{x}\right]\right)$ is (where [.] denotes greatest integer function) (A) 199 (B) 198 (C) 197 (D) None of these

Short answer

Answer: (A) 199

Step-by-step solution

Apply limit properties

We can split the given function into two separate limits, as follows: \(\lim_{x \rightarrow 0} \left[\frac{100 x}{\sin x}\right] + \lim_{x \rightarrow 0} \left[\frac{99 \sin x}{x}\right]\)

Solve the first limit using L'Hôpital's Rule

Since the first limit is of the form \(\frac{0}{0}\) when \(x \to 0\), we can use L'Hôpital's Rule. This states that if \(\lim_{x \rightarrow a} \frac{f(x)}{g(x)}\) exists and is not of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then \(\lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}\) Applying L'Hôpital's Rule to the first limit, we have: \(\lim_{x \rightarrow 0} \frac{100 x}{\sin x} = \lim_{x \rightarrow 0} \frac{100}{\cos x} = \frac{100}{\cos 0} = 100\) So, \([\lim_{x \rightarrow 0} \frac{100 x}{\sin x}] = [100] = 100\).

Solve the second limit using trigonometric identities

Now let's consider the second limit: \(\lim_{x \rightarrow 0} \left[\frac{99 \sin x}{x}\right]\). We can use the trigonometric identity \(\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1\). Thus, \(\lim_{x \rightarrow 0} \left[\frac{99 \sin x}{x}\right] = 99[\lim_{x \rightarrow 0} \frac{\sin x}{x}] = 99[1] = 99\)

Combine the results and choose the correct option

Adding the results from Steps 2 and 3, we get: \([100] + [99] = 100 + 99 = 199\) So, the correct answer is (A) 199.