\(\lim_{{x \to \frac{1}{\sqrt{2}}}} \frac{\sin(\cos^{-1}(x)) - x}{1 - \tan(\cos^{-1}(x))}\)
is equal to :
\(\lim_{{x \to \frac{1}{\sqrt{2}}}} \frac{\sin(\cos^{-1}(x)) - x}{1 - \tan(\cos^{-1}(x))}\)
is equal to :
\(\sqrt2\)
\(-\sqrt2\)
\(\frac{1}{\sqrt2}\)
\(-\frac{1}{\sqrt2}\)
The Correct Option is D
Solution and Explanation
The correct answer is (D) : \(-\frac{1}{\sqrt2}\)
\(\lim_{{x \to \frac{1}{\sqrt{2}}}} \frac{\sin(\cos^{-1}(x)) - x}{1 - \tan(\cos^{-1}(x))}\)
let \(cos^{−1}x=\frac{π}{4}+θ\)
\(\lim_{{\theta \to 0}} \frac{{\sin\left(\frac{\pi}{4} + \theta\right) - \cos\left(\frac{\pi}{4} + \theta\right)}}{{1 - \tan\left(\frac{\pi}{4} + \theta\right)}}\)
\(\lim_{{\theta \to 0}} \frac{{\sqrt{2}\sin\left(\frac{\pi}{4} + \theta - \frac{\pi}{4}\right)}}{{1 - \frac{1 + \tan\theta}{1 - \tan\theta}}}\)
\(\lim_{{\theta \to 0}} \frac{{\sqrt{2}\sin(\theta)}}{{-2\tan(\theta)}}(1 - \tan(\theta) = -\frac{1}{\sqrt{2}}\)
Top Questions on Limits
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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).