Question
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: x→alimtan(x−a)logx−loga?
(a) a1
(b) a
(c) 0
(d) a2
Solution
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 x→alim(f(x)×g(x))=x→alim(f(x))×x→alim(g(x)), x→alim(f(x)1)=x→alim(f(x))1 to proceed through the problem. We then make use of the results x→0lim(xlog(x+a)−loga)=a1 and x→0lim(xtanx)=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 x→alimtan(x−a)logx−loga.
Let us assume L=x→alimtan(x−a)logx−loga ---(1).
Let us assume y=x−a so, we get x=y+a ---(2).
We have given that x→a which leads us to x−a→0⇔y→0 ---(3).
Let us substitute equations (2) and (3) in equation (1).
So, we get L=y→0limtanylog(y+a)−loga ---(4).
Let us multiply the numerator and denominator of the limit with ‘y’ in equation (4).
⇒L=y→0limtany×y(log(y+a)−loga)×y.
⇒L=y→0limy(log(y+a)−loga)×tanyy.
⇒L=y→0limy(log(y+a)−loga)×ytany1.
We know that x→alim(f(x)×g(x))=x→alim(f(x))×x→alim(g(x)).
⇒L=y→0limy(log(y+a)−loga)×y→0limytany1.
We know that x→alim(f(x)1)=x→alim(f(x))1.
⇒L=y→0limy(log(y+a)−loga)×y→0lim(ytany)1.
We know that x→0lim(xlog(x+a)−loga)=a1 and x→0lim(xtanx)=1.
⇒L=a1×11.
⇒L=a1.
So, we have found the value of the limit x→alimtan(x−a)logx−loga as a1.
∴ 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=x→alimtan(x−a)logx−loga.
⇒L=tan(a−a)loga−loga=tan(0)0=00form.
We know that whenever we get the limit in the form of undetermined forms of 00, ∞∞. We can make use of the L-Hospital rule. We know that L-Hospital rule is defined as x→alimg(x)f(x)=x→alimg′(x)f′(x)=x→alimg′′(x)f′′(x)=............
So, we get L=x→alimdxd(tan(x−a))dxd(logx−loga).
⇒L=x→alimdxd(tan(x−a))dxd(logx)−dxd(loga).
⇒L=x→alimsec2(x−a)x1−0.
⇒L=x→alimsec2(x−a)x1.
⇒L=sec2(a−a)a1.
⇒L=sec2(0)a1.
⇒L=1a1.