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Let ƒ : R $$ \to $$ R be a differentiable function satisfying ƒ'(3) + ƒ'(2) = 0. Then $$\mathop {\lim }\limits_{x \to 0} {\left( {{{1 + f(3 + x) - f(3)} \over {1 + f(2 - x) - f(2)}}} \right)^{{1 \over x}}}$$ is equal to

JEE · Math · previous-year question

  1. A.e
  2. B.e2
  3. C.e–1
  4. D.1correct

Answer

D. 1

Explanation

The general formula for indeterminate form 1$$\infty $$ is $$\mathop {\lim }\limits_{x \to a} {\left( {f\left( x \right)} \right)^{g\left( x \right)}} = {e^{\mathop {\lim }\limits_{x \to a} g\left( x \right)\left( {f\left( x \right) - 1} \right)}}$$ I = $$\mathop {\lim }\limits_{x \to 0} {\left( {{{1 + f(3 + x) - f(3)} \over {1 + f(2 - x) - f(2)}}} \right)^{{1 \over x}}}$$ Here I is in 1$$\infty $$ form. $$ \therefore $$ I = $${e^{\mathop {\lim }\limits_{x \to 0} \left( {{{1 + f\left( {3 + x} \right) - f\left( 3 \right)} \over {1 + f\left( {2 - x} \right) - f\left( 2 \right)}} - 1} \right){1 \over x}}}$$ = $${e^{\mathop {\lim }\limits_{x \to 0} \left( {{{1 + f\left( {3 + x} \right) - f\left( 3 \right) - 1 - f\left( {2 - x} \right) + f\left( 2 \right)} \over {1 + f\left( {2 - x} \right) - f\left( 2 \right)}}} \right){1 \over x}}}$$ = $${e^{\mathop {\lim }\limits_{x \to 0} \left( {{{f\left( {3 + x} \right) - f\left( 3 \right) - f\left( {2 - x} \right) + f\left( 2 \right)} \over {1 + f\left( {2 - x} \right) - f\left( 2 \right)}}} \right){1 \over x}}}$$ Here $${{{f\left( {3 + x} \right) - f\left( 3 \right) - f\left( {2 - x} \right) + f\left( 2 \right)} \over x}}$$ is in $${0 \over 0}$$ form. So using L'Hopital rule we get = $${e^{\mathop {\lim }\limits_{x \to 0} \left( {{{f'\left( {3 + x} \right) + f'\left( {2 - x} \right)} \over 1}} \right).\mathop {\lim }\limits_{x \to 0} \left( {{1 \over {1 + f\left( {2 - x} \right) - f\left( 2 \right)}}} \right)}}$$ = $${e^{\left( {f'\left( 3 \right) + f'\left( 2 \right)} \right).1}}$$ = e0 [ as given ƒ'(3) + ƒ'(2) = 0 ] = 1

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