lf `lim_(x to 0) (sin x)/( tan 3x) =a, lim_( x to oo) (sinx)/x =b , lim_( x to oo)( log x)/x = c` then value of a + b + c is

Options

A. 1
B. 1/3
C. 3
D. 4

Answer

To solve the problem, we need to evaluate three limits and then find the sum of their results. Let's break it down step by step.

### Step 1: Evaluate \( a = \lim_{x \to 0} \frac{\sin x}{\tan 3x} \)

We start with the limit:

\[ a = \lim_{x \to 0} \frac{\sin x}{\tan 3x} \]

Recall that \( \tan 3x = \frac{\sin 3x}{\cos 3x} \). Thus, we can rewrite the limit as:

\[ a = \lim_{x \to 0} \frac{\sin x}{\frac{\sin 3x}{\cos 3x}} = \lim_{x \to 0} \frac{\sin x \cdot \cos 3x}{\sin 3x} \]

Now, we can use the fact that \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \). We will manipulate the limit:

\[ a = \lim_{x \to 0} \left( \frac{\sin x}{x} \cdot \frac{x}{\sin 3x} \cdot \cos 3x \right) \]

Now, we know:

\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]

Next, we need to evaluate \( \lim_{x \to 0} \frac{x}{\sin 3x} \):

\[ \lim_{x \to 0} \frac{x}{\sin 3x} = \lim_{x \to 0} \frac{1}{3} \cdot \frac{3x}{\sin 3x} = \frac{1}{3} \cdot 1 = \frac{1}{3} \]

And, since \( \cos 3x \) approaches \( \cos(0) = 1 \) as \( x \to 0 \):

Putting it all together:

\[ a = 1 \cdot \frac{1}{3} \cdot 1 = \frac{1}{3} \]

### Step 2: Evaluate \( b = \lim_{x \to \infty} \frac{\sin x}{x} \)

Now we evaluate:

\[ b = \lim_{x \to \infty} \frac{\sin x}{x} \]

The sine function oscillates between -1 and 1, so:

\[ -\frac{1}{x} \leq \frac{\sin x}{x} \leq \frac{1}{x} \]

As \( x \to \infty \), both bounds approach 0. By the Squeeze Theorem:

\[ b = 0 \]

### Step 3: Evaluate \( c = \lim_{x \to \infty} \frac{\log x}{x} \)

Now we evaluate:

\[ c = \lim_{x \to \infty} \frac{\log x}{x} \]

This is an indeterminate form \( \frac{\infty}{\infty} \), so we can apply L'Hôpital's Rule:

\[ c = \lim_{x \to \infty} \frac{\frac{d}{dx}(\log x)}{\frac{d}{dx}(x)} = \lim_{x \to \infty} \frac{\frac{1}{x}}{1} = \lim_{x \to \infty} \frac{1}{x} = 0 \]

### Step 4: Combine the results

Now we have:

\[ a = \frac{1}{3}, \quad b = 0, \quad c = 0 \]

Thus, we find:

\[ a + b + c = \frac{1}{3} + 0 + 0 = \frac{1}{3} \]

### Final Answer

The value of \( a + b + c \) is:

\[ \boxed{\frac{1}{3}} \]