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Limits Continuity and Differentiability question

2021 · 27 Jul · Shift 2 · Q31
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  5. /2021 · 27 Jul · Shift 2 · Q31

Limits Continuity and Differentiability question

2021 · 27 Jul · Shift 2 · Q31

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
The value of lim⁡x→0(x1−sin⁡x8−1+sin⁡x8)\mathop {\lim }\limits_{x \to 0} \left( {{x \over {\sqrt[8]{1 - \sin x} - \sqrt[8]{1 + \sin x} }}} \right)x→0lim​(81−sinx​−81+sinx​x​) is equal to :
  1. A
    0
  2. B
    4
  3. C
    −-− 4
  4. D
    −-− 1
View written solutionFree

Correct answer: C

  1. We need to evaluate L=lim⁡x→0x1−sin⁡x8−1+sin⁡x8.L=\lim_{x\to 0}\frac{x}{\sqrt[8]{1-\sin x}-\sqrt[8]{1+\sin x}}.L=limx→0​81−sinx​−81+sinx​x​.

  2. As x→0x\to 0x→0, we have sin⁡x→0\sin x\to 0sinx→0, so both denominator terms approach 111. Hence the denominator tends to 1−1=0,1-1=0,1−1=0, and numerator also tends to 000. So this is a 00\frac{0}{0}00​ form.

  3. Let f(t)=(1+t)1/8.f(t)=(1+t)^{1/8}.f(t)=(1+t)1/8. Then the denominator becomes (1−sin⁡x)1/8−(1+sin⁡x)1/8=f(−sin⁡x)−f(sin⁡x).(1-\sin x)^{1/8}-(1+\sin x)^{1/8}=f(-\sin x)-f(\sin x).(1−sinx)1/8−(1+sinx)1/8=f(−sinx)−f(sinx).

For small ttt, using linear approximation, f(t)=f(0)+f′(0)t+⋯f(t)=f(0)+f'(0)t+\cdotsf(t)=f(0)+f′(0)t+⋯ with f′(t)=18(1+t)−7/8  ⟹  f′(0)=18.f'(t)=\frac{1}{8}(1+t)^{-7/8} \implies f'(0)=\frac{1}{8}.f′(t)=81​(1+t)−7/8⟹f′(0)=81​. So, f(t)≈1+t8.f(t)\approx 1+\frac{t}{8}.f(t)≈1+8t​.

Therefore,

\qquad (1+\sin x)^{1/8}\approx 1+\frac{\sin x}{8}.$$ Thus the denominator is approximately $$\left(1-\frac{\sin x}{8}\right)-\left(1+\frac{\sin x}{8}\right)=-\frac{\sin x}{4}.$$ 4. Hence, $$L=\lim_{x\to 0}\frac{x}{\sqrt[8]{1-\sin x}-\sqrt[8]{1+\sin x}} \approx \lim_{x\to 0}\frac{x}{-\frac{\sin x}{4}} =\lim_{x\to 0}-4\frac{x}{\sin x}.$$ 5. Using $$\lim_{x\to 0}\frac{\sin x}{x}=1 \implies \lim_{x\to 0}\frac{x}{\sin x}=1,$$ we get $$L=-4.$$ 6. Therefore the correct option is $$\boxed{\text{C: }-4}.$$
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