# Evaluate the following limit:

$\underset{x\to 2}{\mathop{\lim }}\,\dfrac{x-2}{{{\log }_{a}}(x-1)}$

Last updated date: 21st Mar 2023

•

Total views: 308.4k

•

Views today: 8.86k

Answer

Verified

308.4k+ views

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.

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.

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE