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Limits Continuity and Differentiability question

2019 · 8 Apr · Shift 2 · Q33
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  5. /2019 · 8 Apr · Shift 2 · Q33

Limits Continuity and Differentiability question

2019 · 8 Apr · Shift 2 · Q33

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
Let ƒ : R →\to→ R be a differentiable function satisfying ƒ'(3) + ƒ'(2) = 0. Then lim⁡x→0(1+f(3+x)−f(3)1+f(2−x)−f(2))1x\mathop {\lim }\limits_{x \to 0} {\left( {{{1 + f(3 + x) - f(3)} \over {1 + f(2 - x) - f(2)}}} \right)^{{1 \over x}}}x→0lim​(1+f(2−x)−f(2)1+f(3+x)−f(3)​)x1​ is equal to
  1. A
    e
  2. B
    e2
  3. C
    e–1
  4. D
    1
View written solutionFree

Correct answer: D

  1. Write the limit in a standard exponential form

Let L=lim⁡x→0(1+f(3+x)−f(3)1+f(2−x)−f(2))1/x.L=\lim_{x\to 0}\left(\frac{1+f(3+x)-f(3)}{1+f(2-x)-f(2)}\right)^{1/x}.L=limx→0​(1+f(2−x)−f(2)1+f(3+x)−f(3)​)1/x.

This is of the type 1∞1^{\infty}1∞, so take logarithm: ln⁡L=lim⁡x→01xln⁡(1+f(3+x)−f(3)1+f(2−x)−f(2)).\ln L=\lim_{x\to 0}\frac{1}{x}\ln\left(\frac{1+f(3+x)-f(3)}{1+f(2-x)-f(2)}\right).lnL=limx→0​x1​ln(1+f(2−x)−f(2)1+f(3+x)−f(3)​).

Using log properties, ln⁡L=lim⁡x→01x[ln⁡(1+f(3+x)−f(3))−ln⁡(1+f(2−x)−f(2))].\ln L=\lim_{x\to 0}\frac{1}{x}\left[\ln\big(1+f(3+x)-f(3)\big)-\ln\big(1+f(2-x)-f(2)\big)\right].lnL=limx→0​x1​[ln(1+f(3+x)−f(3))−ln(1+f(2−x)−f(2))].

  1. Use differentiability to expand the small increments

Since fff is differentiable, f(3+x)−f(3)=f′(3)x+o(x),f(3+x)-f(3)=f'(3)x+o(x),f(3+x)−f(3)=f′(3)x+o(x), f(2−x)−f(2)=f′(2)(−x)+o(x)=−f′(2)x+o(x).f(2-x)-f(2)=f'(2)(-x)+o(x)=-f'(2)x+o(x).f(2−x)−f(2)=f′(2)(−x)+o(x)=−f′(2)x+o(x).

So, 1+f(3+x)−f(3)=1+f′(3)x+o(x),1+f(3+x)-f(3)=1+f'(3)x+o(x),1+f(3+x)−f(3)=1+f′(3)x+o(x), 1+f(2−x)−f(2)=1−f′(2)x+o(x).1+f(2-x)-f(2)=1-f'(2)x+o(x).1+f(2−x)−f(2)=1−f′(2)x+o(x).

Hence,

=1+f′(3)x+o(x)1−f′(2)x+o(x).=\frac{1+f'(3)x+o(x)}{1-f'(2)x+o(x)}.=1−f′(2)x+o(x)1+f′(3)x+o(x)​.

Now use 1+ax+o(x)1+bx+o(x)=1+(a−b)x+o(x).\frac{1+ax+o(x)}{1+bx+o(x)}=1+(a-b)x+o(x).1+bx+o(x)1+ax+o(x)​=1+(a−b)x+o(x).

Here b=−f′(2)b=-f'(2)b=−f′(2), so

=1+\big(f'(3)+f'(2)\big)x+o(x).$$ Given $$f'(3)+f'(2)=0,$$ we get $$\frac{1+f(3+x)-f(3)}{1+f(2-x)-f(2)}=1+o(x).$$ Therefore, $$\left(\frac{1+f(3+x)-f(3)}{1+f(2-x)-f(2)}\right)^{1/x} =\big(1+o(x)\big)^{1/x}.$$ --- 3. **Evaluate using logarithm** From above, $$\ln L=\lim_{x\to 0}\frac{1}{x}\ln(1+o(x)).$$ Since $\ln(1+u)\sim u$ as $u\to 0$, and here $u=o(x)$, $$\ln(1+o(x))=o(x).$$ Thus, $$\ln L=\lim_{x\to 0}\frac{o(x)}{x}=0.$$ So, $$L=e^0=1.$$ --- 4. **Check options** - A: $e$ - B: $e^2$ - C: $e^{-1}$ - D: $1$ Hence the correct option is $$\boxed{\text{D}}.$$
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