Question

How do you find the limit of $\dfrac{\dfrac{1}{x+4}-\left( \dfrac{1}{4} \right)}{x}$ as x approaches to zero $\left( x \to 0 \right)$?

Answer 1

Hint: We start solving the problem by performing the subtraction operation to the functions present inside the given limit. We then make the necessary calculations and then take the cancel the common factors present in both numerator and denominator present inside the limit. We then make use of the fact that $\displaystyle \lim_{x \to a}\dfrac{1}{f\left( x \right)+1}=\dfrac{1}{f\left( a \right)+1}$ and then make the necessary calculations to get the required answer.

Complete step by step answer:
According to the problem, we are asked to find the limit of $\dfrac{\dfrac{1}{x+4}-\left( \dfrac{1}{4} \right)}{x}$ as x approaches zero $\left( x \to 0 \right)$.
Let us assume $L=\displaystyle \lim_{x \to 0}\dfrac{\dfrac{1}{x+4}-\left( \dfrac{1}{4} \right)}{x}$.
Let us subtraction the given functions inside the limit.
$\Rightarrow L=\displaystyle \lim_{x \to 0}\dfrac{\dfrac{4-\left( x+4 \right)}{4\left( x+4 \right)}}{x}$.
$\Rightarrow L=\displaystyle \lim_{x \to 0}\dfrac{\dfrac{4-x-4}{4\left( x+4 \right)}}{x}$.
$\Rightarrow L=\displaystyle \lim_{x \to 0}\dfrac{\dfrac{-x}{4\left( x+4 \right)}}{x}$.
We can see that the numerator and denominator inside the limit has a common factor $x$.
$\Rightarrow L=\displaystyle \lim_{x \to 0}\dfrac{-1}{4\left( x+4 \right)}$.
We know that $\displaystyle \lim_{x \to a}\dfrac{1}{f\left( x \right)+1}=\dfrac{1}{f\left( a \right)+1}$.
$\Rightarrow L=\dfrac{-1}{4\left( 0+4 \right)}$.
$\Rightarrow L=\dfrac{-1}{4\left( 4 \right)}$.
\[\Rightarrow L=\dfrac{-1}{16}\].
We have found the value of the limit of $\dfrac{\dfrac{1}{x+4}-\left( \dfrac{1}{4} \right)}{x}$ when x approaches to zero $\left( x \to 0 \right)$ as $\dfrac{-1}{16}$.
$\therefore $ The value of the limit of $\dfrac{\dfrac{1}{x+4}-\left( \dfrac{1}{4} \right)}{x}$ when x approaches to zero $\left( x \to 0 \right)$ is $\dfrac{-1}{16}$.

Note:
We can see that the limit results out to be indeterminate form $\dfrac{0}{0}$ if we substitute the limit 0 in place of x so, we can make use of L-Hospital rule to get the value of given limit. We should not make calculation mistakes while solving this type of problem. Whenever we get this type of problem, we first simplify the function given inside the limit and substitute the value of the given limit of x to get the required answer. Similarly, we can expect problems to find the limit of the given function when x approaches $\infty $.