Let f : R to R be a differentiable function s | vedclass

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(B) The expression is of the form $1^\infty$ as $x 	o 0$.Let $L = \mathop {\lim }\limits_{x 	o 0} {\left( {\frac{{1 + f\left( {3 + x} ight) - f\le

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(B) The expression is of the form $1^\infty$ as $x 	o 0$.Let $L = \mathop {\lim }\limits_{x 	o 0} {\left( {\frac{{1 + f\left( {3 + x} ight) - f\left( 3 ight)}}{{1 + f\left( {2 - x} ight) - f\left( 2 ight)}}} ight)^{\frac{1}{x}}}$.Using the formula $\mathop {\lim }\limits_{x 	o a} {\left( {f(x)} ight)^{g(x)}} = e^{\mathop {\lim }\limits_{x 	o a} g(x)(f(x) - 1)}$,we get:$L = e^{\mathop {\lim }\limits_{x 	o 0} \frac{1}{x} \left( \frac{1 + f(3+x) - f(3)}{1 + f(2-x) - f(2)} - 1 ight)}$$L = e^{\mathop {\lim }\limits_{x 	o 0} \frac{1}{x} \left( \frac{f(3+x) - f(3) - f(2-x) + f(2)}{1 + f(2-x) - f(2)} ight)}$Applying $L$'Hopital's rule to the exponent:Exponent $= \mathop {\lim }\limits_{x 	o 0} \frac{f'(3+x) + f'(2-x)}{1 + f(2-x) - f(2) - xf'(2-x)} = \frac{f'(3) + f'(2)}{1} = 0$.Thus,$L = e^0 = 1$.

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

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