Question

If we have a function $f\left( x \right)$ is $\dfrac{{{10}^{x}}-{{5}^{x}}-{{2}^{x}}+1}{{{x}^{2}}}$, $x\ne 0$ then for f to be continuous at $x=0$, $f\left( 0 \right)$ should be
(a) $\log 10$
(b) $\log 5.\log 2$
(c) $\log {}^{5}/{}_{2}$
(d) none of these

Answer 1

In order to solve these types of question, one must understand if a function is continuous at $x=a$, then $\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( n \right)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( n \right)=f\left( a \right)$. Here, the given function changes to indeterminate form of $\dfrac{0}{0}$ form, if we simply replace the value of x with ‘zero’ in order to find $f\left( 0 \right)$. In order to find $f\left( 0 \right)$, we need to factorize the numerator and denominator, such that to eliminate the indeterminate form. On factorizing, we can use the formula $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{a}^{x}}-1}{x}={{\log }_{e}}a$. This formula will help to eliminate the indeterminant form and obtain the value of $f\left( 0 \right)$.

Complete step-by-step solution:
We have given function as $f\left( x \right)$ which is equal to $\dfrac{{{10}^{x}}-{{5}^{x}}-{{2}^{x}}+1}{{{x}^{2}}}$, $x\ne 0$ . Now, we have to find the value of $f\left( 0 \right)$ for f to be continuous at $x=0$.
In order to make f to be continuous at $x=0$, the value of $f\left( 0 \right)$ will be

$f\left( 0 \right)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{10}^{x}}-{{5}^{x}}-{{2}^{x}}+1}{{{x}^{2}}}\text{ }\left[ \dfrac{0}{0}\text{ form} \right]$
Now, we can factorize the numerator in order to eliminate the indeterminate form, as shown below:
$\Rightarrow f\left( 0 \right)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{5}^{x}}\times {{2}^{x}}-{{5}^{x}}-{{2}^{x}}+1}{{{x}^{2}}}$
Then, we can take common at the numerator, we get,
$\Rightarrow f\left( 0 \right)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{5}^{x}}\left( {{2}^{x}}-1 \right)-\left( {{2}^{x}}-1 \right)}{{{x}^{2}}}$
Again, we common factors at the numerator, so taking the common factors, we get,
$\Rightarrow f\left( 0 \right)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left( {{5}^{x}}-1 \right)\left( {{2}^{x}}-1 \right)}{{{x}^{2}}}$
Then, we can divide the denominator in two terms as below, we get,
$\Rightarrow f\left( 0 \right)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left( {{5}^{x}}-1 \right)}{x}\times \dfrac{\left( {{2}^{x}}-1 \right)}{x}...................(i)$
This function changes to indeterminate if we replace the value of x with ‘zero’. Hence, we can’t get the value of $f\left( 0 \right)$ in this way.
Here, we can make the use of the formula given by $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{a}^{x}}-1}{x}={{\log }_{e}}a...............(ii)$
From equation (i) and (ii), we can clearly see that equation (i) can be converted in the form of equation (ii) using the formula.
So, using equation (i), we get following function from equation (i),
$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{5}^{x}}-1}{x}={{\log }_{e}}5................(iii)$
Similarly, we can also use equation (ii) for other term present in equation (i) as above, hence, we get,
$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{2}^{x}}-1}{x}={{\log }_{e}}2...................(iv)$
Hence, using the equation (i), (iii) and (iv) and replacing the respective values, we get,
$\therefore f\left( 0 \right)={{\log }_{e}}5.{{\log }_{e}}2$
Hence, for f to be continuous at $x=0$, $f\left( 0 \right)$ should be ${{\log }_{e}}5.{{\log }_{e}}2$.
Thus, the correct option is an option (b).

Note: Students need to be clear of the fact that if a function is continuous at $x=a$, then $\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( n \right)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( n \right)=f\left( a \right)$. Students often makes mistake by directly replacing the value of x with ‘zero’ in the given function and declaring the indeterminate form of $\dfrac{0}{0}$ as an answer. Besides, students should remember the formula like $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{a}^{x}}-1}{x}={{\log }_{e}}a$ as these types of formula are very useful in eliminating or changing the indeterminate form to determinate form.