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Limits Continuity and Differentiability question

2022 · 26 Jun · Shift 1 · Q25

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Limits Continuity and Differentiability
2022 · 26 Jun · Shift 1 · Q25

Limits Continuity and Differentiability question

2022 · 26 Jun · Shift 1 · Q25

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1

\mathop {\lim }\limits_{x \to {1 \over {\sqrt 2 }}} {{\sin ({{\cos }^{ - 1}}x) - x} \over {1 - \tan ({{\cos }^{ - 1}}x)}}

is equal to :

A
$\sqrt 2$
B
$- \sqrt 2$
C
${1 \over {\sqrt 2 }}$
D
$- {1 \over {\sqrt 2 }}$

View written solutionFree

Correct answer: D

Rewrite the inverse-trigonometric expressions

Let $\theta = \cos^{-1} x.$ Then $\cos\theta = x, \qquad \theta \in [0,\pi].$

So, $\sin(\cos^{-1}x)=\sin\theta=\sqrt{1-x^2}$ because for $\theta\in[0,\pi]$ , $\sin\theta\ge 0$ .

Also, $\tan(\cos^{-1}x)=\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{\sqrt{1-x^2}}{x}.$

Hence the limit becomes $L=\lim_{x\to \frac1{\sqrt2}} \frac{\sqrt{1-x^2}-x}{1-\frac{\sqrt{1-x^2}}{x}}.$

Simplify the denominator

$1-\frac{\sqrt{1-x^2}}{x}=\frac{x-\sqrt{1-x^2}}{x}.$

Therefore, $L=\lim_{x\to \frac1{\sqrt2}} \frac{\sqrt{1-x^2}-x}{\frac{x-\sqrt{1-x^2}}{x}}.$

Notice that $\sqrt{1-x^2}-x=-(x-\sqrt{1-x^2}).$

So, $L=\lim_{x\to \frac1{\sqrt2}} \frac{-(x-\sqrt{1-x^2})}{\frac{x-\sqrt{1-x^2}}{x}}.$

For $x\neq \frac1{\sqrt2}$ near the limit point, we can cancel $x-\sqrt{1-x^2}$ : $L=\lim_{x\to \frac1{\sqrt2}} (-x).$

Thus, $L=-\frac1{\sqrt2}.$

Check with options

The value is $-\frac1{\sqrt2},$ which corresponds to Option D.

Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.

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