Let f:R\rightarrow R be a differentiable function satisfying f'(3)+f'(2)=0. Then \lim_{x\rightarrow 0}\left ( \frac{1+f(3+x)-f(3)}{1+f(2-x)-f(2)} \right )^{\frac{1}{x}}is equal to:

  • Option 1)

    1

  • Option 2)

    e^{-1}

  • Option 3)

    e

  • Option 4)

    e^{2}

 

Answers (1)

\lim_{x\rightarrow 0}\left ( \frac{1+f(3+x)-f(3)}{1+f(2-x)-f(2)} \right )^{\frac{1}{x}}

1^{\infty} form.

e^{\lim_{x\rightarrow 0}\frac{1}{x}\left ( \frac{1+f(3+x)-f(3)}{1+f(2-x)-f(2)}-1 \right )}

\Rightarrow e^{\lim_{x\rightarrow 0}\frac{1}{x}\left ( \frac{f(3+x)-f(3)-f(2-x)+f(2)}{1+f(2-x)-f(2)} \right )}

\frac{0}{0} form, Apply L'Hapital rule 

\Rightarrow e^{\lim_{x\rightarrow 0} \frac{f'(3+x)+f'(2-x)}{(-f'(2-x)x+f(2-x)-f(2)+1)}}

\Rightarrow e^{f'(3)+f'(2)}=e^{0}=1


Option 1)

1

Option 2)

e^{-1}

Option 3)

e

Option 4)

e^{2}

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