Answer :
To evaluate the limit [tex]\(\lim_{x \to 0} \frac{\operatorname{cosec} x - \cot x}{x}\)[/tex], let's begin by rewriting the trigonometric functions in terms of sine and cosine.
Recall that:
[tex]\[ \operatorname{cosec} x = \frac{1}{\sin x} \quad \text{and} \quad \cot x = \frac{\cos x}{\sin x} \][/tex]
Thus, the given expression becomes:
[tex]\[ \frac{\operatorname{cosec} x - \cot x}{x} = \frac{\frac{1}{\sin x} - \frac{\cos x}{\sin x}}{x} \][/tex]
Combining the terms in the numerator over a common denominator, [tex]\(\sin x\)[/tex], gives:
[tex]\[ \frac{1 - \cos x}{x \sin x} \][/tex]
Next, let's rewrite this as the product of two separate limits:
[tex]\[ \frac{1 - \cos x}{x \sin x} = \frac{1 - \cos x}{x} \cdot \frac{1}{\sin x} \][/tex]
We now evaluate each factor separately.
First, consider the limit:
[tex]\[ \lim_{x \to 0} \frac{1 - \cos x}{x} \][/tex]
We know from the standard limit results that:
[tex]\[ \lim_{x \to 0} \frac{1 - \cos x}{x} = 0 \quad \text{(since } 1 - \cos x \text{ approaches 0 faster than } x \text{ as } x \to 0\text{)} \][/tex]
However, for clarity, let's use a Taylor series expansion for [tex]\(\cos x\)[/tex] around [tex]\(x = 0\)[/tex]:
[tex]\[ \cos x \approx 1 - \frac{x^2}{2} \][/tex]
Then, substituting this approximation:
[tex]\[ 1 - \cos x \approx 1 - \left(1 - \frac{x^2}{2}\right) = \frac{x^2}{2} \][/tex]
Thus,
[tex]\[ \lim_{x \to 0} \frac{1 - \cos x}{x} = \lim_{x \to 0} \frac{\frac{x^2}{2}}{x} = \lim_{x \to 0} \frac{x}{2} = 0 \][/tex]
However, if we consider:
[tex]\[ \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} \][/tex]
This similarity gives a clue when combined properly with more critical conditions for precise solving. Next, consider:
[tex]\[ \lim_{x \to 0} \frac{1}{\sin x} \][/tex]
Using the well-known limit:
[tex]\[ \sin x \approx x \text{ as } x \to 0 \Rightarrow \lim_{x \to 0} \frac{1}{\sin x} = \frac{1}{x} \][/tex]
Combining these:
[tex]\[ \lim_{x \to 0} \left( \frac{1 - \cos x}{x} \cdot \frac{1}{\sin x} \right) = \frac{1}{2} \][/tex]
Therefore:
[tex]\[ \lim_{x \to 0} \frac{\operatorname{cosec} x - \cot x}{x} = \frac{1}{2} \][/tex]
Recall that:
[tex]\[ \operatorname{cosec} x = \frac{1}{\sin x} \quad \text{and} \quad \cot x = \frac{\cos x}{\sin x} \][/tex]
Thus, the given expression becomes:
[tex]\[ \frac{\operatorname{cosec} x - \cot x}{x} = \frac{\frac{1}{\sin x} - \frac{\cos x}{\sin x}}{x} \][/tex]
Combining the terms in the numerator over a common denominator, [tex]\(\sin x\)[/tex], gives:
[tex]\[ \frac{1 - \cos x}{x \sin x} \][/tex]
Next, let's rewrite this as the product of two separate limits:
[tex]\[ \frac{1 - \cos x}{x \sin x} = \frac{1 - \cos x}{x} \cdot \frac{1}{\sin x} \][/tex]
We now evaluate each factor separately.
First, consider the limit:
[tex]\[ \lim_{x \to 0} \frac{1 - \cos x}{x} \][/tex]
We know from the standard limit results that:
[tex]\[ \lim_{x \to 0} \frac{1 - \cos x}{x} = 0 \quad \text{(since } 1 - \cos x \text{ approaches 0 faster than } x \text{ as } x \to 0\text{)} \][/tex]
However, for clarity, let's use a Taylor series expansion for [tex]\(\cos x\)[/tex] around [tex]\(x = 0\)[/tex]:
[tex]\[ \cos x \approx 1 - \frac{x^2}{2} \][/tex]
Then, substituting this approximation:
[tex]\[ 1 - \cos x \approx 1 - \left(1 - \frac{x^2}{2}\right) = \frac{x^2}{2} \][/tex]
Thus,
[tex]\[ \lim_{x \to 0} \frac{1 - \cos x}{x} = \lim_{x \to 0} \frac{\frac{x^2}{2}}{x} = \lim_{x \to 0} \frac{x}{2} = 0 \][/tex]
However, if we consider:
[tex]\[ \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} \][/tex]
This similarity gives a clue when combined properly with more critical conditions for precise solving. Next, consider:
[tex]\[ \lim_{x \to 0} \frac{1}{\sin x} \][/tex]
Using the well-known limit:
[tex]\[ \sin x \approx x \text{ as } x \to 0 \Rightarrow \lim_{x \to 0} \frac{1}{\sin x} = \frac{1}{x} \][/tex]
Combining these:
[tex]\[ \lim_{x \to 0} \left( \frac{1 - \cos x}{x} \cdot \frac{1}{\sin x} \right) = \frac{1}{2} \][/tex]
Therefore:
[tex]\[ \lim_{x \to 0} \frac{\operatorname{cosec} x - \cot x}{x} = \frac{1}{2} \][/tex]