Question:
$ \lim_ {{x \to 0}} \frac { x.2^x-x} {1-cosx} $ is equal to
$ \lim_ {{x \to 0}} \frac { x.2^x-x} {1-cosx} $ is equal to
Updated On: Jul 6, 2022
- log 2
- $\frac {1} {2 } log 2$
- 2 log 2
- 44563
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The Correct Option is C
Solution and Explanation
$\lim_{x\to0} \frac{x-2^{x} -x}{1 -\cos x} = \lim _{x\to 0} \frac{x\left(2^{x} -1\right)}{1-\cos x} $
= $ \lim _{x\to 0} \frac{x^{2}}{1 -\cos x} . \frac{2^{x} -1}{x} $
= $\lim _{x\to 0} \frac{x^{2} }{1-\cos x}.\lim _{x\to 0} \frac{2^{x} -1}{x}$
= $ \lim _{x\to 0} 2\left(\frac{x/2}{\sin x/2}\right)^{2} . \log2 = 2 \log 2$
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Concepts Used:
Limits
A function's limit is a number that a function reaches when its independent variable comes to a certain value. The value (say a) to which the function f(x) approaches casually as the independent variable x approaches casually a given value "A" denoted as f(x) = A.
If limx→a- f(x) is the expected value of f when x = a, given the values of ‘f’ near x to the left of ‘a’. This value is also called the left-hand limit of ‘f’ at a.
If limx→a+ f(x) is the expected value of f when x = a, given the values of ‘f’ near x to the right of ‘a’. This value is also called the right-hand limit of f(x) at a.
If the right-hand and left-hand limits concur, then it is referred to as a common value as the limit of f(x) at x = a and denote it by lim x→a f(x).