Question

How do you evaluate the limit \[\lim \dfrac{{{3}^{x}}-{{2}^{x}}}{x}\] as \[x \to 0\].

Answer 1

Hint: In this problem, we have to evaluate the given limit \[\lim \dfrac{{{3}^{x}}-{{2}^{x}}}{x}\] as \[x \to 0\]. We know that we cannot directly apply the \[x \to 0\] in the limit as the value becomes indeterminate. So, we can use the L ’Hospital Rule, that differentiates numerator and denominator separately, until indeterminate forms exist and then we can substitute \[x \to 0\], to get the answer.

Complete step by step answer:
We know that the given limit to be evaluated is,
\[\displaystyle \lim_{x \to 0}\dfrac{{{3}^{x}}-{{2}^{x}}}{x}\]
 Now we can apply \[x \to 0\] in the above limit, we get
\[\Rightarrow \displaystyle \lim_{x \to 0}\dfrac{{{3}^{x}}-{{2}^{x}}}{x}=\dfrac{{{3}^{0}}-{{2}^{0}}}{0}=\dfrac{0}{0}\].
We know that the above step is in indeterminate form.
We know, L ‘Hospital Rule states that, when the limit of \[\dfrac{f\left( x \right)}{g\left( x \right)}\] is indeterminant, under a certain condition it can be obtained by evaluating the limit of quotient of the derivatives of f and g, i.e., \[\dfrac{f'\left( x \right)}{g'\left( x \right)}\]. If this result is indeterminate, the procedure can be repeated.
Now we can apply the L’ Hospital Rule and differentiate the numerator and denominator separately for the given limit.
\[\begin{align}
  & \Rightarrow \lim \dfrac{{{3}^{x}}-{{2}^{x}}}{x} \\
 & \Rightarrow \lim \dfrac{{{3}^{x}}.\ln 3-{{2}^{x}}.\ln 2}{1} \\
\end{align}\]
Now we can apply the limit in the above step, we get
\[\begin{align}
  & \Rightarrow \displaystyle \lim_{x \to 0}\dfrac{{{3}^{x}}.\ln 3-{{2}^{x}}.\ln 2}{1}=\dfrac{{{3}^{0}}.\ln 3-{{2}^{0}}.\ln 2}{1} \\
 & \Rightarrow \ln 3-\ln 2 \\
 & \Rightarrow \ln \left( \dfrac{3}{2} \right)\text{ }\because \text{ln}3-\ln 2=\ln \left( \dfrac{3}{2} \right) \\
\end{align}\]

Therefore, by evaluating \[\displaystyle \lim_{x \to 0}\dfrac{{{3}^{x}}-{{2}^{x}}}{x}\], the answer is \[\ln \left( \dfrac{3}{2} \right)\].

Note: Students make mistakes while substituting the limit value correctly which should be concentrated. We should know that, if we apply the limit value directly and get the indeterminate form, we should use the L ‘Hospital rule and differentiate both the numerator and denominator individually until the limit becomes an indeterminate form.