Question

Find the value of \[\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\left( {ab} \right)}^x} - {a^x} - {b^x} + 1}}{{{x^2}}} = \]
A) $\log a\log b$
B) $\log \left( {ab} \right)$
C) \[\log \left( {\dfrac{a}{b}} \right)\]
D) $\log \left( {\dfrac{b}{a}} \right)$

Answer 1

Hint:
We can apply the limits directly. If we get the limit as $\dfrac{0}{0}$, we can apply L’ hospital’s rule. So, we can take the derivative of the numerator and denominator. Then we can again apply the limits and apply L’ hospital’s rule if we get the limit as $\dfrac{0}{0}$. Then we can take the derivative of the numerator and denominator and apply the limits to get the required limit.

Complete step by step solution:
We need to find \[\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\left( {ab} \right)}^x} - {a^x} - {b^x} + 1}}{{{x^2}}}\]
Let \[L = \mathop {\lim }\limits_{x \to 0} \dfrac{{{{\left( {ab} \right)}^x} - {a^x} - {b^x} + 1}}{{{x^2}}}\]
On applying the limits, we get
 \[ \Rightarrow L = \dfrac{{{{\left( {ab} \right)}^0} - {a^0} - {b^0} + 1}}{{{0^2}}}\]
We know that ${a^0} = 1$, so we get
 \[ \Rightarrow L = \dfrac{{1 - 1 - 1 + 1}}{0}\]
On simplification, we get
 \[ \Rightarrow L = \dfrac{0}{0}\]
As the limit is $\dfrac{0}{0}$, we can apply L’ hospital’s rule.
We know that by L’ hospital’s rule, If $\mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \dfrac{0}{0}$, then $\mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}$
Here, we have $f\left( x \right) = {\left( {ab} \right)^x} - {a^x} - {b^x} + 1$ and $g\left( x \right) = {x^2}$
Now we can take the derivatives.
 $ \Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\left[ {{{\left( {ab} \right)}^x} - {a^x} - {b^x} + 1} \right]$
We know that $\dfrac{d}{{dx}}{a^x} = {a^x}\log a$. So, we get
 $ \Rightarrow f'\left( x \right) = {\left( {ab} \right)^x}\log ab - {a^x}\log a - {b^x}\log b + 0$
Similarly, we can write
 $ \Rightarrow g'\left( x \right) = 2x$
So, the limit will become,
 $ \Rightarrow L = \mathop {\lim }\limits_{x \to 0} \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}$
On substituting the values, we get
 \[ \Rightarrow L = \mathop {\lim }\limits_{x \to 0} \dfrac{{{{\left( {ab} \right)}^x}\log ab - {a^x}\log a - {b^x}\log b}}{{2x}}\]
On applying the limits, we get
 \[ \Rightarrow L = \dfrac{{{{\left( {ab} \right)}^0}\log ab - {a^0}\log a - {b^0}\log b}}{{2 \times 0}}\]
We know that ${a^0} = 1$, so we get
 \[ \Rightarrow L = \dfrac{{\log ab - \log a - \log b}}{{2 \times 0}}\]
We know that $\log ab = \log a + \log b$, so we get
 \[ \Rightarrow L = \dfrac{{\log a + \log b - \log a - \log b}}{{2 \times 0}}\]
On simplification, we get
 \[ \Rightarrow L = \dfrac{0}{0}\]
Again, the limit is $\dfrac{0}{0}$. So, we can apply L’ hospital’s rule.
Here, we have $f\left( x \right) = {\left( {ab} \right)^x}\log ab - {a^x}\log a - {b^x}\log b$ and $g\left( x \right) = 2x$
Now we can take the derivatives.
 $ \Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\left[ {{{\left( {ab} \right)}^x}\log ab - {a^x}\log a - {b^x}\log b} \right]$
We know that $\dfrac{d}{{dx}}{a^x} = {a^x}\log a$. So, we get
 \[ \Rightarrow f'\left( x \right) = {\left( {ab} \right)^x}{\left( {\log ab} \right)^2} - {a^x}{\left( {\log a} \right)^2} - {b^x}{\left( {\log b} \right)^2}\]
Similarly, we can write
 $ \Rightarrow g'\left( x \right) = 2$
So, the limit will become
 $ \Rightarrow L = \mathop {\lim }\limits_{x \to 0} \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}$
On substituting the values, we get
 \[ \Rightarrow L = \mathop {\lim }\limits_{x \to 0} \dfrac{{{{\left( {ab} \right)}^x}{{\left( {\log ab} \right)}^2} - {a^x}{{\left( {\log a} \right)}^2} - {b^x}{{\left( {\log b} \right)}^2}}}{2}\]
On applying the limits, we get
 \[ \Rightarrow L = \dfrac{{{{\left( {ab} \right)}^0}{{\left( {\log ab} \right)}^2} - {a^0}{{\left( {\log a} \right)}^2} - {b^0}{{\left( {\log b} \right)}^2}}}{2}\]
We know that ${a^0} = 1$, so we get
 \[ \Rightarrow L = \dfrac{{{{\left( {\log ab} \right)}^2} - {{\left( {\log a} \right)}^2} - {{\left( {\log b} \right)}^2}}}{2}\]
We know that $\log ab = \log a + \log b$, so we get
 \[ \Rightarrow L = \dfrac{{{{\left( {\log a + \log b} \right)}^2} - {{\left( {\log a} \right)}^2} - {{\left( {\log b} \right)}^2}}}{2}\]
On expanding the square using the identity ${\left( {x + y} \right)^2} = {x^2} + 2xy + {y^2}$, we get
 \[ \Rightarrow L = \dfrac{{{{\left( {\log a} \right)}^2} + {{\left( {\log b} \right)}^2} + 2\log a\log b - {{\left( {\log a} \right)}^2} - {{\left( {\log b} \right)}^2}}}{2}\]
On simplification, we get
 \[ \Rightarrow L = \dfrac{{2\log a\log b}}{2}\]
On cancelling the common terms, we get
 \[ \Rightarrow L = \log a\log b\]
Therefore, the required limit is \[\log a\log b\].

So, the correct answer is option A.

Note:
We can use only the L’ hospital’s rule when the limit is of the form $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$ . We must identify the functions of numerator and denominator and find their derivative separately. We cannot expand the log of the sum of 2 numbers as the sum of their logarithms. We must use brackets wherever needed to avoid errors. We cannot take the \[{2^{nd}}\] derivative directly. We must check the limit after taking the \[{1^{st}}\] derivative before taking the \[{2^{nd}}\] derivative.