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Question: Compute the limit given in the problem: \(\displaystyle \lim_{x \to a}\dfrac{\log x-\log a}{\tan \le...

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

Explanation

Solution

We start solving the problem by assuming the given limit equal to L. We then assume y=xay=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 limxa(f(x)×g(x))=limxa(f(x))×limxa(g(x))\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), limxa(1f(x))=1limxa(f(x))\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 limx0(log(x+a)logax)=1a\displaystyle \lim_{x \to 0}\left( \dfrac{\log \left( x+a \right)-\log a}{x} \right)=\dfrac{1}{a} and limx0(tanxx)=1\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 limxalogxlogatan(xa)\displaystyle \lim_{x \to a}\dfrac{\log x-\log a}{\tan \left( x-a \right)}.
Let us assume L=limxalogxlogatan(xa)L=\displaystyle \lim_{x \to a}\dfrac{\log x-\log a}{\tan \left( x-a \right)} ---(1).
Let us assume y=xay=x-a so, we get x=y+ax=y+a ---(2).
We have given that xax \to a which leads us to xa0y0x-a\to 0\Leftrightarrow y\to 0 ---(3).
Let us substitute equations (2) and (3) in equation (1).
So, we get L=limy0log(y+a)logatanyL=\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).
L=limy0(log(y+a)loga)×ytany×y\Rightarrow L=\displaystyle \lim_{y\to 0}\dfrac{\left( \log \left( y+a \right)-\log a \right)\times y}{\tan y\times y}.
L=limy0(log(y+a)loga)y×ytany\Rightarrow L=\displaystyle \lim_{y\to 0}\dfrac{\left( \log \left( y+a \right)-\log a \right)}{y}\times \dfrac{y}{\tan y}.
L=limy0(log(y+a)loga)y×1tanyy\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 limxa(f(x)×g(x))=limxa(f(x))×limxa(g(x))\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).
L=limy0(log(y+a)loga)y×limy01tanyy\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 limxa(1f(x))=1limxa(f(x))\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)}.
L=limy0(log(y+a)loga)y×1limy0(tanyy)\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 limx0(log(x+a)logax)=1a\displaystyle \lim_{x \to 0}\left( \dfrac{\log \left( x+a \right)-\log a}{x} \right)=\dfrac{1}{a} and limx0(tanxx)=1\displaystyle \lim_{x \to 0}\left( \dfrac{\tan x}{x} \right)=1.
L=1a×11\Rightarrow L=\dfrac{1}{a}\times \dfrac{1}{1}.
L=1a\Rightarrow L=\dfrac{1}{a}.
So, we have found the value of the limit limxalogxlogatan(xa)\displaystyle \lim_{x \to a}\dfrac{\log x-\log a}{\tan \left( x-a \right)} as 1a\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=limxalogxlogatan(xa)L=\displaystyle \lim_{x \to a}\dfrac{\log x-\log a}{\tan \left( x-a \right)}.
L=logalogatan(aa)=0tan(0)=00form\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 00\dfrac{0}{0}, \dfrac{\infty }{\infty }. We can make use of the L-Hospital rule. We know that L-Hospital rule is defined as limxaf(x)g(x)=limxaf(x)g(x)=limxaf(x)g(x)=...........\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=limxaddx(logxloga)ddx(tan(xa))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)}.
L=limxaddx(logx)ddx(loga)ddx(tan(xa))\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)}.
L=limxa1x0sec2(xa)\Rightarrow L=\displaystyle \lim_{x \to a}\dfrac{\dfrac{1}{x}-0}{{{\sec }^{2}}\left( x-a \right)}.
L=limxa1xsec2(xa)\Rightarrow L=\displaystyle \lim_{x \to a}\dfrac{\dfrac{1}{x}}{{{\sec }^{2}}\left( x-a \right)}.
L=1asec2(aa)\Rightarrow L=\dfrac{\dfrac{1}{a}}{{{\sec }^{2}}\left( a-a \right)}.
L=1asec2(0)\Rightarrow L=\dfrac{\dfrac{1}{a}}{{{\sec }^{2}}\left( 0 \right)}.
L=1a1\Rightarrow L=\dfrac{\dfrac{1}{a}}{1}.