Question

Determine the value for the limit:- $\mathop {\lim }\limits_{x \to 0} \dfrac{{{2^{3x}} - {3^x}}}{{\sin 3x}}$

Answer 1

According to the question, we have to find the value of $\mathop {\lim }\limits_{x \to 0} \dfrac{{{2^{3x}} - {3^x}}}{{\sin 3x}}$.
So, first of all we have to use the L hospital’s rule because as we see that at $\mathop {\lim }\limits_{x \to 0}$ the numerator and denominator of the given question equals to zero.
So, first of all we have to understand the L'hospital's rule with the help of the example which is explained below.
L'hospital's rule: This rule tells us that if we have an intermediate form 0/0 or $\infty /\infty$ all we need to do is differentiate the numerator and differentiate the denominator and then take the limit.
For example: Suppose that we have one of the following cases,
$\mathop { \Rightarrow \lim }\limits_{x \to a} \dfrac{{f(x)}}{{g(x)}} = \dfrac{0}{0}$ OR $\mathop {\lim }\limits_{x \to a} \dfrac{{f(x)}}{{g(x)}} = \dfrac{{ \pm \infty }}{{ \pm \infty }}$
Where $a$ can be any real number, infinity or negative infinity. In these cases we have to differentiate the numerator and differentiate the denominator both and then take the limit.

Complete step-by-step answer:
Step 1: First of all we have to check that at $\mathop {\lim }\limits_{x \to 0}$ the numerator and denominator of the given question equals zero or not.
$\mathop { \Rightarrow \lim }\limits_{x \to 0} \dfrac{{{2^{3x}} - {3^x}}}{{\sin 3x}} \\\ \Rightarrow \dfrac{{{2^{3(0)}} - {3^0}}}{{\sin 3(0)}} \\\ \Rightarrow \dfrac{{{2^0} - {3^0}}}{{\sin (0)}} \\\$
As we know that, if any number has power to 0 then its value equal to 1 and $\sin (0) = 0$
$\Rightarrow \dfrac{{1 - 1}}{0} \\\ \Rightarrow \dfrac{0}{0}.....................................(1) \\\$
Step 2: As we see that in expression (1) in the solution step 1 the functions are in the form 0/0.
Now, we have to use the L hospital’s rule which is mentioned in the solution hint.
$\mathop { \Rightarrow \lim }\limits_{x \to 0} \dfrac{{{2^{3x}} - {3^x}}}{{\sin 3x}} \\\ \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{{8^x} - {3^x}}}{{\sin 3x}}. \\\ \Rightarrow .\mathop { \Rightarrow \lim }\limits_{x \to 0} \dfrac{{{2^{3x}} - {3^x}}}{{\sin 3x}} \\\ \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{d}{{dx}}({8^x}) - \dfrac{d}{{dx}}({3^x})}}{{\dfrac{d}{{dx}}(\sin 3x)}}........................(2) \\\$

As we know that the differentiation of ${a^x}$ is ${a^x}{\log _e}a$ and differentiation of $\sin ax$is $a\cos (ax)$
Step 3: So, now we have to use the differentiation rules which are mentioned just above in the expression (2).
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{{8^x}{{\log }_e}8 - {3^x}{{\log }_e}3}}{{3\cos 3x}}$
Now, put $\mathop {\lim }\limits_{x \to 0}$in the just above expression.
$\Rightarrow \dfrac{{{8^0}{{\log }_e}8 - {3^0}{{\log }_e}3}}{{3\cos 3(0)}}$
As we know that, if any number has power to 0 then its value equal to 1 and $\cos (0) = 1$

$\Rightarrow \dfrac{{(1){{\log }_e}8 - (1){{\log }_e}3}}{{3(1)}} \\\ \Rightarrow \dfrac{{{{\log }_e}8 - {{\log }_e}3}}{3}....................(3) \\\ \\\$
Step 4: Now, we know that$\log m - \log n = \log \dfrac{m}{n}$. So, apply this rule in the expression (3) of solution step (3).
$\Rightarrow \dfrac{{{{\log }_e}8 - {{\log }_e}3}}{3} \\\ \Rightarrow \dfrac{1}{3}\log \dfrac{8}{3} \\\$

As we know that $\dfrac{1}{a}\log \dfrac{m}{n} = \log {\left( {\dfrac{m}{n}} \right)^{\dfrac{1}{a}}}$
$\Rightarrow \log {\left( {\dfrac{8}{3}} \right)^{\dfrac{1}{3}}}$

Hence, the value for $\mathop {\lim }\limits_{x \to 0} \dfrac{{{2^{3x}} - {3^x}}}{{\sin 3x}}$ is $\log {\left( {\dfrac{8}{3}} \right)^{\dfrac{1}{3}}}$.

Note:
It is necessary to check at $\mathop {\lim }\limits_{x \to 0}$ for the given expression $\mathop {\lim }\limits_{x \to 0} \dfrac{{{2^{3x}} - {3^x}}}{{\sin 3x}}$ has numerator and denominator of the given expression equals to zero or not. It is necessary to apply the L hospital’s rule when the given expression has both numerator and denominator in the form of $\dfrac{0}{0}.$