Question

Find the limit of the following:
(1) \[\mathop {\lim }\limits_{x \to 2} \dfrac{{\sqrt {3 - x} - 1}}{{2 - x}}\]
(2) \[\mathop {\lim }\limits_{x \to 0} \dfrac{{{3^{2x}} - {2^{3x}}}}{x}\]

Answer 1

Hint: To solve this, we will first substitute the limit directly, then we see the given expression will be in the form of \[\dfrac{0}{0}\] .So, we will use L-Hospital’s Rule to solve the limit which states that differentiate both numerator and denominator separately. After that we will apply the limit and find the required result. We keep on differentiating till we get a value.

Complete step-by-step answer:
1. The given expression is, \[\mathop {\lim }\limits_{x \to 2} \dfrac{{\sqrt {3 - x} - 1}}{{2 - x}}\]
and we have to find the limit of the given expression,
So, first of all when we substitutes \[x = 2\] in the above expression, then we get the expression in the form of \[\dfrac{0}{0}\] as,
\[\mathop {\lim }\limits_{x \to 2} \dfrac{{\sqrt {3 - x} - 1}}{{2 - x}} = \dfrac{{\sqrt {3 - 2} - 1}}{{2 - 2}} = \dfrac{0}{0}\]
which is an indeterminate form
Now, as we know that if a limit is in the form of \[\dfrac{0}{0}\] then we will use L-Hospital’s Rule to find out the limit.
So, by applying L-Hospital’s Rule on the given expression, we get
\[\mathop {\lim }\limits_{x \to 2} \dfrac{{\dfrac{d}{{dx}}\left( {\sqrt {3 - x} - 1} \right)}}{{\dfrac{d}{{dx}}\left( {2 - x} \right)}}\]
Each algebraic expression in both numerator and denominator is formed by the difference of two expressions. The derivative of the difference of the terms can be evaluated by the difference of their derivatives as per the difference rule of the derivatives.
\[ \Rightarrow \mathop {\lim }\limits_{x \to 2} \dfrac{{\dfrac{d}{{dx}}\left( {\sqrt {3 - x} } \right) - \dfrac{d}{{dx}}\left( 1 \right)}}{{\dfrac{d}{{dx}}\left( 2 \right) - \dfrac{d}{{dx}}\left( x \right)}}\]
Now, we know that \[\dfrac{d}{{dx}}\left( c \right) = 0\] where c is constant,
And \[\dfrac{d}{{dx}}\left( {\sqrt x } \right) = \dfrac{{ - 1}}{{2\sqrt x }}\]
Therefore, on differentiating above expression, we get
\[ \Rightarrow \mathop {\lim }\limits_{x \to 2} \dfrac{{\dfrac{{ - 1}}{{2\sqrt {3 - x} }}}}{{ - 1}}\]
\[ \Rightarrow \mathop {\lim }\limits_{x \to 2} \dfrac{1}{{2\sqrt {3 - x} }}\]
Now as \[x\] approaches \[2\] ,so substitute the value of \[x = 2\] directly in the above expression, we get
\[\dfrac{1}{{2\sqrt {3 - 2} }}\]
On simplifying, we get
\[ = \dfrac{1}{2}\]
$\therefore $ \[\mathop {\lim }\limits_{x \to 2} \dfrac{{\sqrt {3 - x} - 1}}{{2 - x}} = \dfrac{1}{2}\].

Now, we will solve the second part in the same way,
2. The given expression is, \[\mathop {\lim }\limits_{x \to 0} \dfrac{{{3^{2x}} - {2^{3x}}}}{x}\]
and we have to find the limit of the given expression,
So, first of all when we substitutes \[x = 0\] in the above expression, then we get the expression in the form of \[\dfrac{0}{0}\] as,
\[\mathop {\lim }\limits_{x \to 0} \dfrac{{{3^{2x}} - {2^{3x}}}}{x} = \dfrac{{{3^0} - {2^0}}}{0} = \dfrac{{1 - 1}}{0} = \dfrac{0}{0}\]
which is an indeterminate form
Now, as we know that if a limit is in the form of \[\dfrac{0}{0}\] then we will use L-Hospital’s Rule to find out the limit.
So, by applying L-Hospital’s Rule on the given expression, we get
\[\mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{d}{{dx}}\left( {{3^{2x}} - {2^{3x}}} \right)}}{{\dfrac{d}{{dx}}\left( x \right)}}\]
Each algebraic expression in both numerator is formed by the difference of two expressions. The derivative of the difference of the terms can be evaluated by the difference of their derivatives as per the difference rule of the derivatives.
\[ \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{d}{{dx}}\left( {{3^{2x}}} \right) - \dfrac{d}{{dx}}\left( {{2^{3x}}} \right)}}{{\dfrac{d}{{dx}}\left( x \right)}}\]
Now, we know that \[\dfrac{d}{{dx}}\left( {{a^x}} \right) = {a^x}{\log _e}a\]
Therefore, on differentiating above expression, we get
\[ \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{2\left( {{3^{2x}}{{\log }_e}3} \right) - 3\left( {{2^{3x}}{{\log }_e}2} \right)}}{1}\]
\[ \Rightarrow \mathop {\lim }\limits_{x \to 0} 2\left( {{3^{2x}}{{\log }_e}3} \right) - 3\left( {{2^{3x}}{{\log }_e}2} \right)\]
Now as \[x\] approaches \[0\] ,so substitute the value of \[x = 0\] directly in the above expression, we get
\[2\left( {{3^0}{{\log }_e}3} \right) - 3\left( {{2^0}{{\log }_e}2} \right)\]
On simplifying, we get
\[ = 2{\log _e}3 - 3{\log _e}2\]
Now, we know that \[m{\log _e}a = {\log _e}{a^m}\]
Therefore, we get
\[ = {\log _e}{3^2} - {\log _e}{2^3}\]
\[ = {\log _e}9 - {\log _e}8\]
Now, we know that,
\[{\log _e}m - {\log _e}n = {\log _e}\dfrac{m}{n}\]
\[ \Rightarrow {\log _e}9 - {\log _e}8 = {\log _e}\dfrac{9}{8}\]
$\therefore $ \[\mathop {\lim }\limits_{x \to 0} \dfrac{{{3^{2x}} - {2^{3x}}}}{x} = {\log _e}\dfrac{9}{8}\].

Note: Whenever we face such types of problems the main point to remember is that first of all always use a direct method to determine whether the limit is in determinate or indeterminate form. If it turns out to be in determinate form, then the limit is solved by direct substitution method only. And if the limit turns out to be an indeterminate form, then we use L-Hospital’s Rule to solve further. This helps in getting us the required condition and gets us on the right track to reach the answer.