Problem 250 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \lim _{x \rightarrow 0} \frac{1}{\sin x}-\frac{1}{\tan x}\\{\text { Ans. } 0\\} $$

Step-by-Step Solution

Answer

To find the limit of \(\lim _{x \rightarrow 0} \frac{1}{\sin x}-\frac{1}{\tan x}\), we rewrite the function using a common denominator, simplify the expression, and apply L'Hopital's Rule. After these steps, we find that the limit is 0: \(\lim _{x \rightarrow 0} \frac{1}{\sin x} - \frac{1}{\tan x}=0\).

Step 1: Rewrite the function using a common denominator

We can rewrite the function as a single fraction with a common denominator. The common denominator for sin(x) and tan(x) is sin(x)tan(x): \(\lim _{x \rightarrow 0} \frac{1}{\sin x}-\frac{1}{\tan x} = \lim _{x \rightarrow 0} \frac{\tan(x)-\sin(x)}{\sin(x) \tan(x)}\)

Step 2: Simplify the expression

Now, we can simplify the expression, replacing tan(x) with sin(x)/cos(x): \(\lim _{x \rightarrow 0} \frac{\tan(x)-\sin(x)}{\sin(x) \tan(x)} = \lim_{x \rightarrow 0} \frac{\frac{\sin(x)}{\cos(x)} - \sin(x)}{\sin(x) \cdot \frac{\sin(x)}{\cos(x)}}\)

Step 3: Cancel terms and further simplify

Here, we can cancel the sin(x) terms from both the numerator and denominator: \(\lim_{x \rightarrow 0} \frac{\frac{\sin(x)}{\cos(x)} - \sin(x)}{\sin^{2}(x) / \cos(x)} = \lim_{x \rightarrow 0} \frac{\cos(x) - 1}{\sin^{2}(x)}\)

Step 4: Apply L'Hopital's Rule

Now, we can apply L'Hopital's Rule as we have an indeterminate form of 0/0. We take the derivative of both the numerator and denominator until we get an expression without the indeterminate form: \(\lim_{x \rightarrow 0} \frac{-\sin(x)}{2 \sin(x) \cos(x)}\) As x approaches 0, sin(x) approaches 0. Then the limit becomes: \(\lim_{x \rightarrow 0} \frac{-\sin(x)}{2 \sin(x) \cos(x)} = \lim_{x \rightarrow 0} \frac{-1}{2\cos(x)}\) The limit of this expression is 0 as the sine function approaches 0, and the cosine function approaches 1. So, the final answer is:

Step 5: State the result

\(\lim _{x \rightarrow 0} \frac{1}{\sin x} - \frac{1}{\tan x} = 0\)