Problem 41 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \lim _{x \rightarrow \frac{\pi}{2}} \frac{1}{\ln (\sin x)}\\{\text { Ans. }-\infty\\} $$

Step-by-Step Solution

Answer

As x approaches π/2, the sine function approaches 1, which makes the natural logarithm function approach 0. This causes the function \( \frac{1}{\ln (\sin x)} \) to approach negative infinity as the denominator approaches 0 with a positive value in the numerator. Therefore, the limit is \( \lim_{x \rightarrow \frac{\pi}{2}} \frac{1}{\ln (\sin x)} = -\infty \).

Step 1: Analyze the sine function as x approaches π/2

As x approaches π/2, we need to consider the behavior of the sine function. We know that: \( \sin (\frac{\pi}{2}) = 1 \)

Step 2: Analyze the natural logarithm function as sine approaches 1

Now that we know the sine function approaches 1 as x approaches π/2, we can analyze the behavior of the natural logarithm function: \( \ln (\sin x) \) approaches \( \ln (1) \) as x approaches π/2. We also know that: \( \ln(1) = 0 \)

Step 3: Analyze the reciprocal function as logarithm approaches 0

Finally, we analyze the reciprocal function \( \frac{1}{\ln (\sin x)} \). As \( \ln(\sin x) \) approaches 0 when x approaches π/2, the function will approach the form \( \frac{1}{0} \).

Step 4: Determine the limit as x approaches π/2

Since the function approaches the form \( \frac{1}{0} \) as x approaches π/2, and 1 is a positive value, the function is going to negative infinity when approaching \( \frac{\pi}{2} \) from the left side. Therefore, \( \lim_{x \rightarrow \frac{\pi}{2}} \frac{1}{\ln (\sin x)} = -\infty \)