SATHEE: Limit Continuity and Differentiability 1 Question 14

Limit Continuity and Differentiability 1 Question 14

16. $lim_{x \to 1} \frac{\sqrt{1 - \cos 2 (x - 1)}}{x - 1}$

(1998, 2M)

(a) exists and it equals $\sqrt{2}$

(b) exists and it equals $- \sqrt{2}$

(d) does not exist because left hand limit is not equal to right hand limit

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Answer:

Correct Answer: 16. (c)

Solution:

$lim_{x \to 0} \tan \frac{π}{4} + x$

$\begin{aligned} = lim_{x \to 0} \frac{\tan \frac{π}{4} + \tan x}{1 - \tan \frac{π}{4} \tan x} = lim_{x \to 0} \frac{1 + \tan x}{1 - \tan x} \\ = lim_{x \to 0} \frac{{[(1 + \tan x)^{1 / \tan x}]}^{\tan x / x}}{{[(1 - \tan x)^{- 1 / \tan x}]}^{- \tan x / x}} = \frac{e^{1}}{e^{- 1}} = e^{2} \end{aligned}$

Limit Continuity and Differentiability 1 Question 14

16. $lim_{x \to 1} \frac{\sqrt{1 - \cos 2 (x - 1)}}{x - 1}$

Answer:

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Limit Continuity and Differentiability 1 Question 14

16. limx→11−cos⁡2(x−1)x−1

Answer:

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Medical Preparation

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16. $lim_{x \to 1} \frac{\sqrt{1 - \cos 2 (x - 1)}}{x - 1}$