Question

In the given question, \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\ln (2+x)+\ln 0.5}{x}\] is equal to
A. \[\dfrac{1}{2}\]
B. \[\dfrac{3}{2}\]
C. \[-\dfrac{1}{2}\]
D. 1

Answer 1

Hint: An important property of logarithm is ln (xy)=ln (x)+ln (y).
Another important trick that may be used be used here that is expansion of log (1+x) which is
\[\log (1+x)=x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}+.......\] (For \[ -1 < x \leq 1 \])

Complete step-by-step answer:

As mentioned in the question, we have this expression in which we can write it using the property mentioned in the hint part as
\[\begin{align}
  & \underset{x\to 0}{\mathop{=\lim }}\,\dfrac{\ln (2+x)+\ln 0.5}{x} \\
 & =\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\ln \left( (2+x)0.5 \right)}{x} \\
 & =\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\ln \left( \dfrac{(2+x)}{2} \right)}{x} \\
 & =\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\ln \left( 1+\dfrac{x}{2} \right)}{x} \\
\end{align}\]
Now, using the expansion of the logarithmic function as mentioned in the hint part as
\[\begin{align}
  & =\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\ln \left( 1+\dfrac{x}{2} \right)}{x} \\
 & =\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{x}{2}-\dfrac{{{\left( \dfrac{x}{2} \right)}^{2}}}{2}+\dfrac{{{\left( \dfrac{x}{2} \right)}^{3}}}{3}-\dfrac{{{\left( \dfrac{x}{2} \right)}^{4}}}{4}+.......}{x} \\
 & =\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{x}{2}-\dfrac{{{x}^{2}}}{{{2}^{2}}\cdot 2}+\dfrac{{{x}^{3}}}{{{2}^{3}}\cdot 3}-\dfrac{{{x}^{4}}}{{{2}^{4}}\cdot 4}+.......}{x} \\
 & =\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x}{2\cdot x}-\dfrac{{{x}^{2}}}{{{2}^{2}}\cdot 2\cdot x}+\dfrac{{{x}^{3}}}{{{2}^{3}}\cdot 3\cdot x}-\dfrac{{{x}^{4}}}{{{2}^{4}}\cdot 4\cdot x}+....... \\
 & =\underset{x\to 0}{\mathop{\lim }}\,\dfrac{1}{2}-\dfrac{x}{{{2}^{2}}\cdot 2}+\dfrac{{{x}^{2}}}{{{2}^{3}}\cdot 3}-\dfrac{{{x}^{3}}}{{{2}^{4}}\cdot 4}+.......\ \ \ \ ...(a) \\
 & =\dfrac{1}{2} \\
\end{align}\]
Hence, by putting the value of x in the equation (a), we get that every term except the first one becomes 0 due to the presence of the extra x in each term after the first one.
The only x that was present in the first term of the sequence or the expansion got cancelled from the extra x that was being divided with each term.
Hence, the value of the limit is \[\dfrac{1}{2}\] .

Note: The students can get confused in solving this question if they will not consider the 2nd term of the question that is ln (0.5) into the first term with the use of the properties as mentioned in the hint which are the basic properties of a logarithmic function because the expansion of ln (1+x) will show up only on involving the second term into the first term in the question.