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Limits Continuity and Differentiability question

2019 · 9 Jan · Shift 2 · Q43
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Limits Continuity and Differentiability question

2019 · 9 Jan · Shift 2 · Q43

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
For each x ∈\in∈ R, let [x] be the greatest integer less than or equal to x. Then lim⁡x→0−  x([x]+∣x∣)sin⁡[x]∣x∣\mathop {\lim }\limits_{x \to {0^ - }} \,\,{{x\left( {\left[ x \right] + \left| x \right|} \right)\sin \left[ x \right]} \over {\left| x \right|}}x→0−lim​∣x∣x([x]+∣x∣)sin[x]​ is equal to :
  1. A
    −-− sin 1
  2. B
    1
  3. C
    sin 1
  4. D
    0
View written solutionFree

Correct answer: A

  1. We need to evaluate L=lim⁡x→0−x([x]+∣x∣)sin⁡[x]∣x∣.L=\lim_{x\to 0^-}\frac{x\left([x]+|x|\right)\sin [x]}{|x|}.L=limx→0−​∣x∣x([x]+∣x∣)sin[x]​.

  2. Since the limit is as x→0−x\to 0^-x→0−, we have x<0x<0x<0 and very close to 000.

For such xxx:

  • ∣x∣=−x|x|=-x∣x∣=−x
  • and because −1<x<0-1<x<0−1<x<0, we get [x]=−1.[x]=-1.[x]=−1.
  1. Substitute these into the expression:
=\frac{x\left(-1+|x|\right)\sin(-1)}{|x|}.$$ Now use $|x|=-x$: $$=\frac{x(-1-x)\sin(-1)}{-x}.$$ Since $\dfrac{x}{-x}=-1$ for $x<0$, $$= -(-1-x)\sin(-1)=(1+x)\sin(-1).$$ Also, $\sin(-1)=-\sin 1$, so $$ (1+x)\sin(-1)=-(1+x)\sin 1.$$ 4. Taking the limit as $x\to 0^-$, $$L=-(1+0)\sin 1=-\sin 1.$$ 5. Hence the correct option is $$\boxed{A\; :\; -\sin 1}.$$
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