Question

If $\mathrm{f}(\mathrm{x})=\left\\{\begin{array}{cl}\frac{\sin ([\mathrm{x}]+2 \mathrm{x})}{[\mathrm{x}]} & \text { if }[\mathrm{x}] \neq 0 \\ 0 & \text { if }[\mathrm{x}]=0\end{array}\right.\(, where \)[.]$ denotes the greatest integer function, then \(\lim _{x \rightarrow 0} f(x)\) is (A) 0 (B) 1 (C) \(-1\) (D) none of these

Short answer

(A) The limit exists and equals 0 (B) The limit exists and equals 2 (C) The limit exists and equals 1 (D) None of these Answer: (D) None of these.

Step-by-step solution

Define the given function

We are given the function f(x) in a piece-wise format: $$ f(x)=\begin{cases} \frac{\sin([x]+2x)}{[x]}, & \text{if }[x]\neq 0 \\ 0, & \text{if }[x]=0 \end{cases} $$

Write the limit expression

We need to find the limit of the function f(x) as x approaches 0. We can write the limit expression as: $$ \lim_{x\to 0}f(x) $$

Analyze the limit for different cases

Since the function f(x) is given in a piece-wise defined format, let's break the limit down into two cases, one when [x] = 0 and one when [x] ≠ 0. Case 1: \([x] = 0\) When [x] equals 0, the function f(x) is directly given as 0. So, as x approaches 0 and [x] is 0, we have: $$ \lim_{x\to 0} f(x) = 0 $$ Case 2: \([x] ≠ 0\) Note that when \(x \to 0\), we have \(0<x<1\). So for this case we can consider: $$ f(x) = \frac{\sin([x]+2x)}{[x]} = \frac{\sin(2x)}{[x]} $$ Since \(0<x<1\), [x]=0, hence: $$ \lim_{x\to 0} f(x) = \lim_{x\to 0} \frac{\sin(2x)}{0} $$ This is an indeterminate form (0/0). To resolve this limit, let's use L'Hopital's Rule.

Apply L'Hopital's Rule

L'Hopital's Rule states that for a limit of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), the limit can be found by computing the limit of the derivative of the numerator and the derivative of the denominator separately. We'll apply L'Hopital's Rule to the limit above: $$ \lim_{x\to 0} f(x) = \lim_{x\to 0} \frac{d(\sin(2x))}{d(0)} $$

Compute the derivatives

We compute the derivative of the numerator and skip the constant 0 in the denominator: $$ \frac{d(\sin(2x))}{dx} = 2\cos(2x) $$

Evaluate the limit

Now we can find the limit: $$ \lim_{x\to 0} f(x) = \lim_{x\to 0} 2\cos(2x) = 2\cos(0) = 2 $$ However, this result corresponds to x > 0.

Combine the cases

In summary, we found that the limit of f(x) is: * 0, when x = 0 * 2, when 0 < x < 1 Therefore, as x approaches 0 from different sides, it has different limits. This means that the limit of f(x) does not exist when x tends to 0.

Select the correct answer

Considering the analysis above, the correct answer is: (D) none of these