Question

The following limit
$\underset{x\to 1}{\mathop{\lim }}\,({{\log }_{2}}2x){{\log }_{2}}5$ is equal to :
A. ${{\log }_{2}}5$
B. ${{e}^{{{\log }_{2}}5}}$
C. $e$
D. 0

Answer 1

- Hint: First, check if the Right Hand and Left Hand Limits of this limit exist in the first place, and if they’re equal. If they aren’t equal, or if any of them doesn’t exist, then the limit as a whole won’t exist as a whole. If both the RHL and LHL exist, then the limit exists and is equal to the left and right hand limits

Complete step-by-step solution -
The first step to evaluating any limit is to check whether both, it’s left hand and right-hand limits exist.
This means that we need to check that the limit gives the same value whether we approach $x=1$ from the left side of 1, or from the right side of 1.
Now, we have, Left Hand Limit = $\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,({{\log }_{2}}2x){{\log }_{2}}5$
We know that $\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,f\left( a-h \right)$
Hence, we have
Left Hand Limit
\[\begin{align}
  & =\underset{h\to 0}{\mathop{\lim }}\,\left( {{\log }_{2}}\left( 2\left( 1-h \right) \right){{\log }_{2}}5 \right) \\
 & =\underset{h\to 0}{\mathop{\lim }}\,{{\log }_{2}}\left( 2-2h \right){{\log }_{2}}5={{\log }_{2}}2{{\log }_{2}}5={{\log }_{2}}5 \\
\end{align}\]
Now, for the right hand limit, we’ll make $x$ approach 1 from the right hand side. This means that now, we’re considering $x$ to have a value that is slightly greater than 1.
Now, we have Right Hand Limit = $\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,({{\log }_{2}}2x){{\log }_{2}}5$
We know that $\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,f\left( a+h \right)$
Hence, we have
Left Hand Limit
\[\begin{align}
  & =\underset{h\to 0}{\mathop{\lim }}\,\left( {{\log }_{2}}\left( 2\left( 1+h \right) \right){{\log }_{2}}5 \right) \\
 & =\underset{h\to 0}{\mathop{\lim }}\,{{\log }_{2}}\left( 2+2h \right){{\log }_{2}}5={{\log }_{2}}2{{\log }_{2}}5={{\log }_{2}}5 \\
\end{align}\]
Thus, Left Hand Limit = Right Hand Limit.
This makes us know that since both the LHL and the RHL exist, and are equal, the limit $\underset{x\to 1}{\mathop{\lim }}\,({{\log }_{2}}2x){{\log }_{2}}5$ exists, and is equal to ${{\log }_{2}}5$
Thus, the answer is option A.

Note: It is very important that before evaluating any limit, we check that both its LHL and RHL exist and are equal to each other. If that’s not the case, then the limit’s value is considered to not exist. Thus, if you jump right into substituting the limit’s value into the expression without checking if the limit exists or not, you might reach an answer, but that answer will be the one got from approaching the limit from either the left hand side or the right hand side, and it won’t equal to the answer got by evaluating it when x is made to approach the limiting value from the other side. Hence, your answer will be wrong, since in cases where the limit does not exist, simply substituting gets you a value, but that value is wrong.