Question

lim_(x→0)\((\frac {1+tanx}{1+sinx})^{cosec x}\) = ?

• A. 0
• B. e
• C. 1
• D. \(\frac {1}{e}\)

Answer 1

The Correct Option is C

Let's consider the limit as x approaches 0: 
lim_(x→0) \((\frac {1+tanx}{1+sinx})^{cosec x}\) as x approaches 0 
Substituting x = 0 into the expression gives us: 
\((\frac {1+tan(0)}{1+sin(0)})^{cosec (0)}\)
Since tan(0) = 0 and sin(0) = 0, we have: 
\((\frac {1+0}{1+0})\)^{(cosec 0)} 
Simplifying further: 
1^{(cosec 0)} 
1 
Therefore, the correct option is (C) 1.

Answer 2

\(\lim_{x \to 0} \left( \frac{1 + \tan x}{1 + \sin x} \right)^{\cosec x}\)

we can use approximations and simplifications. Here's a streamlined approach:

Approximate tan x and sin x for small x:
\(\tan x \approx x\)
\(\sin x \approx x\)

Rewrite the expression using these approximations:
\(\frac{1 + \tan x}{1 + \sin x} \approx \frac{1 + x}{1 + x} = 1\)

Consider the entire expression:
\(\left( \frac{1 + \tan x}{1 + \sin x} \right)^{\cosec x} \approx 1^{\cosec x}\)

Note that \(\cosec x = \frac{1}{\sin x}\), and for small x, \(\sin x \approx x\):
\(\cosec x \approx \frac{1}{x}\)

Substitute \(\cosec x \approx \frac{1}{x}\) into the expression:
\(1^{\cosec x} \approx 1^{\frac{1}{x}}\)

Recognize that 1 raised to any power is still 1:
\(1^{\frac{1}{x}} = 1\)

So, the correct option is (C): 1