Question

If we have the limit as $\displaystyle \lim_{x \to 0}\dfrac{\log \left( 3+x \right)-\log \left( 3-x \right)}{x}=k$. Then find the value of k.
A. $-\dfrac{2}{3}$
B. 0
C. $-\dfrac{1}{3}$
D. $\dfrac{2}{3}$

Answer 1

Hint: We first apply the logarithmic theorem of \[\log \left( a \right)-\log \left( b \right)=\log \left( \dfrac{a}{b} \right)\] to find the proper value of x. Then we apply the limit theorem of log for $\displaystyle \lim_{x \to 0}\dfrac{{{\log }_{e}}\left( 1+x \right)}{x}=1$. The change of variable will work as the limit value remains the same. After that we place the value of the new variable to find the solution of the problem.

Complete step-by-step solution
We have the limit theorem of $\displaystyle \lim_{x \to 0}\dfrac{{{\log }_{e}}\left( 1+x \right)}{x}=1$.
For our given limit function, we use limit theorem \[\log \left( a \right)-\log \left( b \right)=\log \left( \dfrac{a}{b} \right)\].
So, from $\log \left( 3+x \right)-\log \left( 3-x \right)$, we get $\log \left( \dfrac{3+x}{3-x} \right)$.
We convert the function to make it similar of ${{\log }_{e}}\left( 1+x \right)$.
${{\log }_{e}}\left( \dfrac{3+x}{3-x} \right)={{\log }_{e}}\left( \dfrac{3-x+2x}{3-x} \right)={{\log }_{e}}\left( 1+\dfrac{2x}{3-x} \right)$.
As it’s given that $x \to 0$, we get $\dfrac{2x}{3-x}\to \dfrac{2\times 0}{3}\to 0$. We take $\dfrac{2x}{3-x}=z$.
So, $\displaystyle \lim_{x \to 0}\dfrac{\log \left( 3+x \right)-\log \left( 3-x \right)}{x}=\displaystyle \lim_{x \to 0}\left[ \dfrac{{{\log }_{e}}\left( 1+z \right)}{z}\times \dfrac{z}{x} \right]$.
We have the theorem that $\displaystyle \lim_{x \to a}\left[ f\left( x \right)g\left( x \right) \right]=\displaystyle \lim_{x \to a}f\left( x \right)\times \displaystyle \lim_{x \to a}g\left( x \right)$.
$\displaystyle \lim_{x \to 0}\left[ \dfrac{{{\log }_{e}}\left( 1+z \right)}{z}\times \dfrac{z}{x} \right]=\left[ \displaystyle \lim_{z\to 0}\dfrac{{{\log }_{e}}\left( 1+z \right)}{z} \right]\times \left[ \displaystyle \lim_{x \to 0}\dfrac{z}{x} \right]=\displaystyle \lim_{x \to 0}\dfrac{z}{x}$.
Even though the function changed its variable, the limit value being the same we have the same result.
Now we place the value of z to get the limit value.
\[\displaystyle \lim_{x \to 0}\dfrac{z}{x}=\displaystyle \lim_{x \to 0}\dfrac{2x}{x\left( 3-x \right)}=\displaystyle \lim_{x \to 0}\dfrac{2}{\left( 3-x \right)}=\dfrac{2}{3}\].
So, the limit value of the function $\displaystyle \lim_{x \to 0}\dfrac{\log \left( 3+x \right)-\log \left( 3-x \right)}{x}=k$ is $\dfrac{2}{3}$.
Therefore, the value of k is $\dfrac{2}{3}$. The correct option is D.

Note: We need to remember when we are applying the limit value the limiting variable comes into work. Other than that, we can keep any limit variable we want. So, the term $\displaystyle \lim_{x \to 0}\dfrac{z}{x}$ is valid even though the variables are mixed.