Show that `Lim_(x to 0 ) (| sin x |)/x ` does not exist .

Answer

To show that the limit \(\lim_{x \to 0} \frac{|\sin x|}{x}\) does not exist, we will evaluate the left-hand limit (LHL) and the right-hand limit (RHL) separately.

### Step 1: Evaluate the Left-Hand Limit (LHL)

The left-hand limit is defined as:
\[ \lim_{x \to 0^-} \frac{|\sin x|}{x} \]
When \(x\) approaches \(0\) from the left (i.e., \(x\) is slightly negative), we have:
\[ |\sin x| = -\sin x \quad \text{(since \(\sin x\) is negative for small negative \(x\))} \]
Thus, we can rewrite the limit as:
\[ \lim_{x \to 0^-} \frac{|\sin x|}{x} = \lim_{x \to 0^-} \frac{-\sin x}{x} \]
We know from the standard limit that:
\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]
Therefore:
\[ \lim_{x \to 0^-} \frac{-\sin x}{x} = -1 \]
So, we have:
\[ \text{LHL} = -1 \]

### Step 2: Evaluate the Right-Hand Limit (RHL)

The right-hand limit is defined as:
\[ \lim_{x \to 0^+} \frac{|\sin x|}{x} \]
When \(x\) approaches \(0\) from the right (i.e., \(x\) is slightly positive), we have:
\[ |\sin x| = \sin x \quad \text{(since \(\sin x\) is positive for small positive \(x\))} \]
Thus, we can rewrite the limit as:
\[ \lim_{x \to 0^+} \frac{|\sin x|}{x} = \lim_{x \to 0^+} \frac{\sin x}{x} \]
Again, using the standard limit:
\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]
So, we have:
\[ \text{RHL} = 1 \]

### Step 3: Conclusion

Since the left-hand limit and the right-hand limit are not equal:
\[ \text{LHL} = -1 \quad \text{and} \quad \text{RHL} = 1 \]
We conclude that:
\[ \lim_{x \to 0} \frac{|\sin x|}{x} \text{ does not exist.} \]