Evaluate the following limit:
$\underset{x\to 2}{\mathop{\lim }}\,\dfrac{x-2}{{{\log }_{a}}(x-1)}$

Hint: First, check if the limit exists by evaluating, both, the Left Hand and the Right Hand Limits. If they
exist, move on and see if this expression has a $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$ form. If it does,
then apply L’ Hopital Rule to find the limit.
Firstly, let’s check if both the Right Hand Limit (RHL) and the Left Hand Limit (LHL) for this limit
exist or not.
For the Left Hand limit, our limit will more appropriately be $\underset{x\to {{2}^{-}}}{\mathop{\lim
}}\,\dfrac{x-2}{{{\log }_{a}}(x-1)}$, since when we are calculating the LHL, we take $x$to be equal to a
value that is slightly less than the limiting value, which, here $=2$.
Thus, evaluating the LHL further, let’s substitute the for $x$ in the limit. Doing so, we get :
$\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,\dfrac{x-2}{{{\log }_{a}}(x-1)}=\dfrac{{{2}^{-}}-2}{{{\log
}_{a}}({{2}^{-}}-1)}=\dfrac{{{0}^{-}}}{{{\log }_{a}}({{1}^{-}})}=\dfrac{{{0}^{-}}}{{{0}^{-}}}=\dfrac{0}{0}$
Thus, the LHL gives us a $\dfrac{0}{0}$ form. Now, let’s see if the RHL evaluates to the same form as
well.
Putting $x$ as a value that is slightly greater than $2$, or putting $x\to {{2}^{+}}$, we’ll get the limit as
$\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,\dfrac{x-2}{{{\log }_{a}}(x-1)}=\dfrac{{{2}^{+}}-2}{{{\log
}_{a}}({{2}^{+}}-1)}=\dfrac{{{0}^{+}}}{{{\log }_{a}}({{1}^{+}})}=\dfrac{{{0}^{+}}}{{{0}^{+}}}=\dfrac{0}{0}$
Thus, we can see that the RHL evaluates to a $\dfrac{0}{0}$ form.
Now, let’s apply the L' Hopital Rule for finding the limit.
L’ Hopital Rule states that any limit which has the $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$ form,
can be found by separately differentiating the numerator and denominator in terms of the variable that
has the limit, until the limit loses its $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$ form.
Mathematically, it states that $\underset{x\to b}{\mathop{\lim }}\,\dfrac{f(x)}{g(x)}=\underset{x\to
b}{\mathop{\lim }}\,\dfrac{f'(x)}{g'(x)}$ till the fraction obtained on successive differentiations loses its
$\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$ form.
Since here, we can see that the limit has a $\dfrac{0}{0}$ form, we can apply L’ Hopital Rule here.
The given equation to evaluate is;
$\underset{x\to 2}{\mathop{\lim }}\,\dfrac{x-2}{{{\log }_{a}}(x-1)}$.
Thus, applying L’ Hopital Rule once, we’ll have \[\begin{align}

& \underset{x\to 2}{\mathop{\lim }}\,\dfrac{x-2}{{{\log }_{a}}(x-1)}=\underset{x\to 2}{\mathop{\lim
}}\,\dfrac{\dfrac{d(x-2)}{dx}}{\dfrac{d({{\log }_{a}}(x-1))}{dx}}=\underset{x\to 2}{\mathop{\lim
}}\,\dfrac{1}{\dfrac{1.\ln (a)}{x-1}} \\
& =\underset{x\to 2}{\mathop{\lim }}\,\dfrac{x-1}{\ln (a)} \\
\end{align}\]
Now, here we can clearly see that the denominator we get after applying L’ Hopital Rule once, is a
constant, since $\ln (a)$will obviously be a constant, owing to $a$being a constant itself.
Thus, we don’t have a $\dfrac{0}{0}$ form anymore, and hence, we can directly substitute the limiting
value of $x$ into the expression we have now. Doing so, we get :
$\underset{x\to 2}{\mathop{\lim }}\,\dfrac{x-1}{\ln (a)}=\dfrac{2-1}{\ln (a)}=\dfrac{1}{\ln
(a)}=\dfrac{1}{{{\log }_{e}}a}={{\log }_{a}}e$
Thus, we finally have the value of the limit, and it is equal to ${{\log }_{a}}e$.
Note: Don’t forget to check the RHL and LHL every time before you opt for a question of limits. It might so
happen, that you apply L’ Hopital Rule and get an answer, but in reality the limit doesn’t exist. Also, be
carefully to stop differentiating as soon as you have a form that’s not indeterminate, while applying L’
Hopital Rule. Chances are, if you go on differentiating, you might get a zero in the denominator, in which
case the expression becomes undefined.