Question

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

Answer 1

Hint: Use the formula $a={{e}^{{{\log }_{e}}a}}$ followed by the use of the property $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\log }_{e}}\left( 1+x \right)}{x}=1$ . 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 $f'\left( 3 \right)+f'\left( 2 \right)=0$ 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 ${{1}^{\infty }}$, which is an indeterminate form.
So, using the formula $a={{e}^{{{\log }_{e}}a}}$ in our expression, we get
$\underset{x\to 0}{\mathop{\lim }}\,{{\left( \dfrac{1+f(3+x)-f(3)}{1+f\left( 2-x \right)-f\left( 2 \right)} \right)}^{\dfrac{1}{x}}}$
$={{e}^{\log {{\left( \underset{x\to 0}{\mathop{\lim }}\,\left( \dfrac{1+f(3+x)-f(3)}{1+f\left( 2-x \right)-f\left( 2 \right)} \right) \right)}^{\dfrac{1}{x}}}}}$
Now, we know that limit inside the log can be taken outside it and $\log {{a}^{b}}=b\log a$ . If we use this in our expression, we get
\[={{e}^{\underset{x\to 0}{\mathop{\lim }}\,\log {{\left( \dfrac{1+f(3+x)-f(3)}{1+f\left( 2-x \right)-f\left( 2 \right)} \right)}^{\dfrac{1}{x}}}}}\]
$={{e}^{\underset{x\to 0}{\mathop{\lim }}\,\text{ }\dfrac{1}{x}{{\log }_{e}}\left( \dfrac{1+f(3+x)-f(3)}{1+f\left( 2-x \right)-f\left( 2 \right)} \right)}}$
Now we will add 1 and subtract 1 from the part inside the log. On doing so, we get
$={{e}^{\underset{x\to 0}{\mathop{\lim }}\,\text{ }\dfrac{1}{x}{{\log }_{e}}\left( \dfrac{1+f(3+x)-f(3)}{1+f\left( 2-x \right)-f\left( 2 \right)}-1+1 \right)}}$
$={{e}^{\underset{x\to 0}{\mathop{\lim }}\,\text{ }\dfrac{1}{x}{{\log }_{e}}\left( \dfrac{1+f(3+x)-f(3)-1-f\left( 2-x \right)+f\left( 2 \right)}{1+f\left( 2-x \right)-f\left( 2 \right)}+1 \right)}}={{e}^{\underset{x\to 0}{\mathop{\lim }}\,\text{ }\dfrac{1}{x}{{\log }_{e}}\left( \dfrac{f(3+x)-f(3)-f\left( 2-x \right)+f\left( 2 \right)}{1+f\left( 2-x \right)-f\left( 2 \right)}+1 \right)}}$
Now we will divide and multiply the numerator and the denominator with the same term and apply the formula $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\log }_{e}}\left( 1+x \right)}{x}=1$, which will give:
$={{e}^{\underset{x\to 0}{\mathop{\lim }}\,\text{ }\dfrac{1}{x}\times \dfrac{f(3+x)-f(3)-f\left( 2-x \right)+f\left( 2 \right)}{1+f\left( 2-x \right)-f\left( 2 \right)}\times \dfrac{1+f\left( 2-x \right)-f\left( 2 \right)}{f(3+x)-f(3)-f\left( 2-x \right)+f\left( 2 \right)}\times {{\log }_{e}}\left( \dfrac{1+f(3+x)-f(3)-1-f\left( 2-x \right)+f\left( 2 \right)}{1+f\left( 2-x \right)-f\left( 2 \right)}+1 \right)}}$
$={{e}^{\underset{x\to 0}{\mathop{\lim }}\,\text{ }\dfrac{1}{x}\left( \dfrac{f(3+x)-f(3)-f\left( 2-x \right)+f\left( 2 \right)}{1+f\left( 2-x \right)-f\left( 2 \right)} \right)}}$
Now as $\dfrac{0}{0}$ form, applying l-hospital’s rule, which states that if a function is of the form $\dfrac{0}{0}$, then differentiate the numerator and the denominator separately to eliminate the indeterminacy. So, we get
\[={{e}^{\underset{x\to 0}{\mathop{\lim }}\,\text{ }}}^{\dfrac{\left( f'(3+x)+f'\left( 2-x \right) \right)\left( 1+f\left( 2-x \right)-f\left( 2 \right) \right)+\left( f(3+x)-f(3)-f\left( 2-x \right)+f\left( 2 \right) \right)f'\left( 2-x \right)}{{{\left( 1+f\left( 2-x \right)-f\left( 2 \right) \right)}^{2}}}}\]
Now on putting the limit, we get
\[={{e}^{\dfrac{0}{1}}}={{e}^{0}}=1\]
Therefore, the answer to the above question is 1.

Note: Whenever you come across the forms $\dfrac{0}{0}$ 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.