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Limits Continuity and Differentiability question

2003 · Shift 0 · Q70
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  5. /2003 · Shift 0 · Q70

Limits Continuity and Differentiability question

2003 · Shift 0 · Q70

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
If lim⁡x→0log⁡(3+x)−log⁡(3−x)x\mathop {\lim }\limits_{x \to 0} {{\log \left( {3 + x} \right) - \log \left( {3 - x} \right)} \over x}x→0lim​xlog(3+x)−log(3−x)​ = k, the value of k is
  1. A
    −23- {2 \over 3}−32​
  2. B
    0
  3. C
    −13- {1 \over 3}−31​
  4. D
    23{2 \over 3}32​
View written solutionFree

Correct answer: D

  1. We need to evaluate
lim⁡x→0log⁡(3+x)−log⁡(3−x)x.\lim_{x\to 0}\frac{\log(3+x)-\log(3-x)}{x}.x→0lim​xlog(3+x)−log(3−x)​.

Let this limit be kkk.

  1. Use the logarithm property:
log⁡(3+x)−log⁡(3−x)=log⁡(3+x3−x).\log(3+x)-\log(3-x)=\log\left(\frac{3+x}{3-x}\right).log(3+x)−log(3−x)=log(3−x3+x​).

So,

k=lim⁡x→0log⁡(3+x3−x)x.k=\lim_{x\to 0}\frac{\log\left(\frac{3+x}{3-x}\right)}{x}.k=x→0lim​xlog(3−x3+x​)​.
  1. A standard way is to use derivatives. Define
f(x)=log⁡(3+x)−log⁡(3−x).f(x)=\log(3+x)-\log(3-x).f(x)=log(3+x)−log(3−x).

Then

k=lim⁡x→0f(x)−f(0)x−0=f′(0),k=\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=f'(0),k=x→0lim​x−0f(x)−f(0)​=f′(0),

since

f(0)=log⁡3−log⁡3=0.f(0)=\log 3-\log 3=0.f(0)=log3−log3=0.
  1. Differentiate f(x)f(x)f(x):
f′(x)=13+x−(−13−x)=13+x+13−x.f'(x)=\frac{1}{3+x}-\left(\frac{-1}{3-x}\right)=\frac{1}{3+x}+\frac{1}{3-x}.f′(x)=3+x1​−(3−x−1​)=3+x1​+3−x1​.

Therefore,

f′(0)=13+13=23.f'(0)=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}.f′(0)=31​+31​=32​.

Hence,

k=23.k=\frac{2}{3}.k=32​.
  1. Check options:
  • A: −23-\frac{2}{3}−32​ — incorrect
  • B: 000 — incorrect
  • C: −13-\frac{1}{3}−31​ — incorrect
  • D: 23\frac{2}{3}32​ — correct

Therefore, the correct answer is Option D.

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