Question

Examine the continuity of the function:
\[f\left( x \right) = \dfrac{{\log 100 + \log \left( {0.01 + x} \right)}}{{3x}},{\text{ for }}x \ne 0\] and $f\left( x \right) = \dfrac{{100}}{3},{\text{ for }}x = 0$; at $x = 0$

Answer 1

Hint: A function is continuous at a point when its left hand limit at that point is equal to its right hand limit at the same point which in turn is equal to the value of the function at that same point. Find out the left hand and right hand limit at $x = 0$ and check if they both are equal to $f(0)$ or not.

Complete step by step answer:
The given function is:
\[f\left( x \right) = \dfrac{{\log 100 + \log \left( {0.01 + x} \right)}}{{3x}},{\text{ for }}x \ne 0\] and \[f\left( x \right) = \dfrac{{100}}{3}{\text{, for }}x \ne 0\]
We know that $\log m + \log n = \log mn$, applying this for $f\left( x \right)$, we’ll get:
\[
   \Rightarrow f\left( x \right) = \dfrac{{\log 100 + \log \left( {0.01 + x} \right)}}{{3x}},{\text{ for }}x \ne 0 \\
   \Rightarrow f\left( x \right) = \dfrac{{\log \left( {1 + 100x} \right)}}{{3x}},{\text{ for }}x \ne 0 \\
\]
If we consider left hand limit, we’ll get:
$ \Rightarrow {\text{L}}{\text{.H}}{\text{.L = }}\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{\log \left( {1 + 100x} \right)}}{{3x}}$,
If we put $x = 0$, we’ll get$\dfrac{0}{0}$ form. So, we will apply L’ Hospital rule i.e. we’ll differentiate numerator and denominator separately. We’ll get:
\[
   \Rightarrow {\text{L}}{\text{.H}}{\text{.L = }}\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{\dfrac{d}{{dx}}\left( {\log \left( {1 + 100x} \right)} \right)}}{{\dfrac{d}{{dx}}\left( {3x} \right)}}, \\
   \Rightarrow {\text{L}}{\text{.H}}{\text{.L = }}\mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{100}}{{\left( {1 + 100x} \right) \times 3}} \\
\]
Now, putting $x = 0$, we’ll get:
\[
   \Rightarrow {\text{L}}{\text{.H}}{\text{.L = }}\dfrac{{100}}{{\left( {1 + 100 \times 0} \right) \times 3}}, \\
   \Rightarrow {\text{L}}{\text{.H}}{\text{.L = }}\dfrac{{100}}{3} \\
\]
Similarly if we consider right hand limit, we’ll get:
$ \Rightarrow {\text{R}}{\text{.H}}{\text{.L = }}\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\log \left( {1 + 100x} \right)}}{{3x}}$
If we put $x = 0$, we’ll get$\dfrac{0}{0}$ form. So, we will again apply the L' Hospital rule. We’ll get:
\[
   \Rightarrow {\text{R}}{\text{.H}}{\text{.L = }}\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\dfrac{d}{{dx}}\left( {\log \left( {1 + 100x} \right)} \right)}}{{\dfrac{d}{{dx}}\left( {3x} \right)}}, \\
   \Rightarrow {\text{R}}{\text{.H}}{\text{.L = }}\mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{100}}{{\left( {1 + 100x} \right) \times 3}} \\
\]
Now, putting $x = 0$, we’ll get:
\[
   \Rightarrow {\text{R}}{\text{.H}}{\text{.L = }}\dfrac{{100}}{{\left( {1 + 100 \times 0} \right) \times 3}}, \\
   \Rightarrow {\text{R}}{\text{.H}}{\text{.L = }}\dfrac{{100}}{3} \\
\]
And from the question, it is given that $f\left( 0 \right) = \dfrac{{100}}{3}$.
So, at $x = 0$, we have:
${\text{L}}{\text{.H}}{\text{.L}} = {\text{R}}{\text{.H}}{\text{.L}} = f\left( x \right)$

Therefore, the given function is continuous.

Note: If a function is not continuous at any point then its differentiation will not exist at that point. Because it will also be non-differentiable at that point. Also a function is continuous at points which are in its domain only.