Question

Compute the limit given in the problem: $\displaystyle \lim_{x \to a}\dfrac{\log x-\log a}{\tan \left( x-a \right)}$?
(a) $\dfrac{1}{a}$
(b) a
(c) 0
(d) $\dfrac{2}{a}$

Answer 1

Hint: We start solving the problem by assuming the given limit equal to L. We then assume $y=x-a$ and find the change in limit with respect to y. We then make all the necessary arrangements in the limit L and then multiply the numerator and denominator with ‘y’. We then make use of the results $\displaystyle \lim_{x \to a}\left( f\left( x \right)\times g\left( x \right) \right)=\displaystyle \lim_{x \to a}\left( f\left( x \right) \right)\times \displaystyle \lim_{x \to a}\left( g\left( x \right) \right)$, $\displaystyle \lim_{x \to a}\left( \dfrac{1}{f\left( x \right)} \right)=\dfrac{1}{\displaystyle \lim_{x \to a}\left( f\left( x \right) \right)}$ to proceed through the problem. We then make use of the results $\displaystyle \lim_{x \to 0}\left( \dfrac{\log \left( x+a \right)-\log a}{x} \right)=\dfrac{1}{a}$ and $\displaystyle \lim_{x \to 0}\left( \dfrac{\tan x}{x} \right)=1$ to get the required value of limit.

Complete step-by-step solution
According to the problem, we are asked to find the given limit $\displaystyle \lim_{x \to a}\dfrac{\log x-\log a}{\tan \left( x-a \right)}$.
Let us assume $L=\displaystyle \lim_{x \to a}\dfrac{\log x-\log a}{\tan \left( x-a \right)}$ ---(1).
Let us assume $y=x-a$ so, we get $x=y+a$ ---(2).
 We have given that $x \to a$ which leads us to $x-a\to 0\Leftrightarrow y\to 0$ ---(3).
Let us substitute equations (2) and (3) in equation (1).
So, we get $L=\displaystyle \lim_{y\to 0}\dfrac{\log \left( y+a \right)-\log a}{\tan y}$ ---(4).
Let us multiply the numerator and denominator of the limit with ‘y’ in equation (4).
$\Rightarrow L=\displaystyle \lim_{y\to 0}\dfrac{\left( \log \left( y+a \right)-\log a \right)\times y}{\tan y\times y}$.
$\Rightarrow L=\displaystyle \lim_{y\to 0}\dfrac{\left( \log \left( y+a \right)-\log a \right)}{y}\times \dfrac{y}{\tan y}$.
$\Rightarrow L=\displaystyle \lim_{y\to 0}\dfrac{\left( \log \left( y+a \right)-\log a \right)}{y}\times \dfrac{1}{\dfrac{\tan y}{y}}$.
We know that $\displaystyle \lim_{x \to a}\left( f\left( x \right)\times g\left( x \right) \right)=\displaystyle \lim_{x \to a}\left( f\left( x \right) \right)\times \displaystyle \lim_{x \to a}\left( g\left( x \right) \right)$.
$\Rightarrow L=\displaystyle \lim_{y\to 0}\dfrac{\left( \log \left( y+a \right)-\log a \right)}{y}\times \displaystyle \lim_{y\to 0}\dfrac{1}{\dfrac{\tan y}{y}}$.
We know that $\displaystyle \lim_{x \to a}\left( \dfrac{1}{f\left( x \right)} \right)=\dfrac{1}{\displaystyle \lim_{x \to a}\left( f\left( x \right) \right)}$.
$\Rightarrow L=\displaystyle \lim_{y\to 0}\dfrac{\left( \log \left( y+a \right)-\log a \right)}{y}\times \dfrac{1}{\displaystyle \lim_{y\to 0}\left( \dfrac{\tan y}{y} \right)}$.
We know that $\displaystyle \lim_{x \to 0}\left( \dfrac{\log \left( x+a \right)-\log a}{x} \right)=\dfrac{1}{a}$ and $\displaystyle \lim_{x \to 0}\left( \dfrac{\tan x}{x} \right)=1$.
$\Rightarrow L=\dfrac{1}{a}\times \dfrac{1}{1}$.
$\Rightarrow L=\dfrac{1}{a}$.
So, we have found the value of the limit $\displaystyle \lim_{x \to a}\dfrac{\log x-\log a}{\tan \left( x-a \right)}$ as $\dfrac{1}{a}$.
$\therefore$ The correct option for the given problem is (a).

Note: We can see that the given problem contains a huge amount of calculation, so we need to perform each step carefully to avoid mistakes. We can also solve this problem as shown below:
So, we have $L=\displaystyle \lim_{x \to a}\dfrac{\log x-\log a}{\tan \left( x-a \right)}$.
$\Rightarrow L=\dfrac{\log a-\log a}{\tan \left( a-a \right)}=\dfrac{0}{\tan \left( 0 \right)}=\dfrac{0}{0}form$.
We know that whenever we get the limit in the form of undetermined forms of $\dfrac{0}{0}$, $\dfrac{\infty }{\infty }$. We can make use of the L-Hospital rule. We know that L-Hospital rule is defined as $\displaystyle \lim_{x \to a}\dfrac{f\left( x \right)}{g\left( x \right)}=\displaystyle \lim_{x \to a}\dfrac{{{f}^{'}}\left( x \right)}{{{g}^{'}}\left( x \right)}=\displaystyle \lim_{x \to a}\dfrac{{{f}^{''}}\left( x \right)}{{{g}^{''}}\left( x \right)}=...........$.
So, we get $L=\displaystyle \lim_{x \to a}\dfrac{\dfrac{d}{dx}\left( \log x-\log a \right)}{\dfrac{d}{dx}\left( \tan \left( x-a \right) \right)}$.
$\Rightarrow L=\displaystyle \lim_{x \to a}\dfrac{\dfrac{d}{dx}\left( \log x \right)-\dfrac{d}{dx}\left( \log a \right)}{\dfrac{d}{dx}\left( \tan \left( x-a \right) \right)}$.
$\Rightarrow L=\displaystyle \lim_{x \to a}\dfrac{\dfrac{1}{x}-0}{{{\sec }^{2}}\left( x-a \right)}$.
$\Rightarrow L=\displaystyle \lim_{x \to a}\dfrac{\dfrac{1}{x}}{{{\sec }^{2}}\left( x-a \right)}$.
$\Rightarrow L=\dfrac{\dfrac{1}{a}}{{{\sec }^{2}}\left( a-a \right)}$.
$\Rightarrow L=\dfrac{\dfrac{1}{a}}{{{\sec }^{2}}\left( 0 \right)}$.
$\Rightarrow L=\dfrac{\dfrac{1}{a}}{1}$.