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Question: Let a function \(f:R\to R\) be a differentiable function satisfying \(f'\left( 3 \right)+f'\left( 2 ...

Let a function f:R→Rf:R\to R be a differentiable function satisfying f′(3)+f′(2)=0f'\left( 3 \right)+f'\left( 2 \right)=0 . Then lim⁡x→0 (1+f(3+x)−f(3)1+f(2−x)−f(2))1x\underset{x\to 0}{\mathop{\lim }}\,{{\left( \dfrac{1+f(3+x)-f(3)}{1+f\left( 2-x \right)-f\left( 2 \right)} \right)}^{\dfrac{1}{x}}} is equal to

Explanation

Solution

Use the formula a=elog⁡eaa={{e}^{{{\log }_{e}}a}} followed by the use of the property lim⁡x→0 log⁡e(1+x)x=1\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\log }_{e}}\left( 1+x \right)}{x}=1 . Then use the l-hospital’s rule to further simplify the resultant expression you get. Once you have eliminated all removable terms, put the limit and use the relation f′(3)+f′(2)=0f'\left( 3 \right)+f'\left( 2 \right)=0 given in the question to reach the answer.

Complete step-by-step answer:
In the expression given in the question, if we put the limit and check, then we will find that the expression turns out to be of the form 1∞{{1}^{\infty }}, which is an indeterminate form.
So, using the formula a=elog⁡eaa={{e}^{{{\log }_{e}}a}} in our expression, we get
lim⁡x→0 (1+f(3+x)−f(3)1+f(2−x)−f(2))1x\underset{x\to 0}{\mathop{\lim }}\,{{\left( \dfrac{1+f(3+x)-f(3)}{1+f\left( 2-x \right)-f\left( 2 \right)} \right)}^{\dfrac{1}{x}}}
=elog⁡(lim⁡x→0 (1+f(3+x)−f(3)1+f(2−x)−f(2)))1x={{e}^{\log {{\left( \underset{x\to 0}{\mathop{\lim }}\,\left( \dfrac{1+f(3+x)-f(3)}{1+f\left( 2-x \right)-f\left( 2 \right)} \right) \right)}^{\dfrac{1}{x}}}}}
Now, we know that limit inside the log can be taken outside it and log⁡ab=blog⁡a\log {{a}^{b}}=b\log a . If we use this in our expression, we get
=elim⁡x→0 log⁡(1+f(3+x)−f(3)1+f(2−x)−f(2))1x={{e}^{\underset{x\to 0}{\mathop{\lim }}\,\log {{\left( \dfrac{1+f(3+x)-f(3)}{1+f\left( 2-x \right)-f\left( 2 \right)} \right)}^{\dfrac{1}{x}}}}}
=elim⁡x→0  1xlog⁡e(1+f(3+x)−f(3)1+f(2−x)−f(2))={{e}^{\underset{x\to 0}{\mathop{\lim }}\,\text{ }\dfrac{1}{x}{{\log }_{e}}\left( \dfrac{1+f(3+x)-f(3)}{1+f\left( 2-x \right)-f\left( 2 \right)} \right)}}
Now we will add 1 and subtract 1 from the part inside the log. On doing so, we get
=elim⁡x→0  1xlog⁡e(1+f(3+x)−f(3)1+f(2−x)−f(2)−1+1)={{e}^{\underset{x\to 0}{\mathop{\lim }}\,\text{ }\dfrac{1}{x}{{\log }_{e}}\left( \dfrac{1+f(3+x)-f(3)}{1+f\left( 2-x \right)-f\left( 2 \right)}-1+1 \right)}}
=elim⁡x→0  1xlog⁡e(1+f(3+x)−f(3)−1−f(2−x)+f(2)1+f(2−x)−f(2)+1)=elim⁡x→0  1xlog⁡e(f(3+x)−f(3)−f(2−x)+f(2)1+f(2−x)−f(2)+1)={{e}^{\underset{x\to 0}{\mathop{\lim }}\,\text{ }\dfrac{1}{x}{{\log }_{e}}\left( \dfrac{1+f(3+x)-f(3)-1-f\left( 2-x \right)+f\left( 2 \right)}{1+f\left( 2-x \right)-f\left( 2 \right)}+1 \right)}}={{e}^{\underset{x\to 0}{\mathop{\lim }}\,\text{ }\dfrac{1}{x}{{\log }_{e}}\left( \dfrac{f(3+x)-f(3)-f\left( 2-x \right)+f\left( 2 \right)}{1+f\left( 2-x \right)-f\left( 2 \right)}+1 \right)}}
Now we will divide and multiply the numerator and the denominator with the same term and apply the formula lim⁡x→0 log⁡e(1+x)x=1\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\log }_{e}}\left( 1+x \right)}{x}=1, which will give:
=elim⁡x→0  1x×f(3+x)−f(3)−f(2−x)+f(2)1+f(2−x)−f(2)×1+f(2−x)−f(2)f(3+x)−f(3)−f(2−x)+f(2)×log⁡e(1+f(3+x)−f(3)−1−f(2−x)+f(2)1+f(2−x)−f(2)+1)={{e}^{\underset{x\to 0}{\mathop{\lim }}\,\text{ }\dfrac{1}{x}\times \dfrac{f(3+x)-f(3)-f\left( 2-x \right)+f\left( 2 \right)}{1+f\left( 2-x \right)-f\left( 2 \right)}\times \dfrac{1+f\left( 2-x \right)-f\left( 2 \right)}{f(3+x)-f(3)-f\left( 2-x \right)+f\left( 2 \right)}\times {{\log }_{e}}\left( \dfrac{1+f(3+x)-f(3)-1-f\left( 2-x \right)+f\left( 2 \right)}{1+f\left( 2-x \right)-f\left( 2 \right)}+1 \right)}}
=elim⁡x→0  1x(f(3+x)−f(3)−f(2−x)+f(2)1+f(2−x)−f(2))={{e}^{\underset{x\to 0}{\mathop{\lim }}\,\text{ }\dfrac{1}{x}\left( \dfrac{f(3+x)-f(3)-f\left( 2-x \right)+f\left( 2 \right)}{1+f\left( 2-x \right)-f\left( 2 \right)} \right)}}
Now as 00\dfrac{0}{0} form, applying l-hospital’s rule, which states that if a function is of the form 00\dfrac{0}{0}, then differentiate the numerator and the denominator separately to eliminate the indeterminacy. So, we get
=elim⁡x→0  (f′(3+x)+f′(2−x))(1+f(2−x)−f(2))+(f(3+x)−f(3)−f(2−x)+f(2))f′(2−x)(1+f(2−x)−f(2))2={{e}^{\underset{x\to 0}{\mathop{\lim }}\,\text{ }}}^{\dfrac{\left( f'(3+x)+f'\left( 2-x \right) \right)\left( 1+f\left( 2-x \right)-f\left( 2 \right) \right)+\left( f(3+x)-f(3)-f\left( 2-x \right)+f\left( 2 \right) \right)f'\left( 2-x \right)}{{{\left( 1+f\left( 2-x \right)-f\left( 2 \right) \right)}^{2}}}}
Now on putting the limit, we get
=e01=e0=1={{e}^{\dfrac{0}{1}}}={{e}^{0}}=1
Therefore, the answer to the above question is 1.

Note: Whenever you come across the forms 00\dfrac{0}{0} or ∞∞\dfrac{\infty }{\infty } always give a try to l-hospital’s rule as this may give you the answer in the fastest possible manner. Also, keep a habit of checking the indeterminate forms of the expressions before starting with the questions of limit as this helps you to select the shortest possible way to reach the answer.