Question

Let $f(x) = \left\{ {\dfrac{{\log {{(1 + x)}^{1 + x}} - x}}{{{x^2}}}} \right\}$ , then find the value of $f(0)$ so that the function is continuous at $x = 0$ .

Answer 1

Hint: In this question we have to find the value of $f(0)$ so that the function is continuous at $x = 0$ .
So we can write this as $f(0) = \mathop {\lim }\limits_{x \to 0} f(x)$ . We know that if we get the expression in the form of $\dfrac{0}{0}$ , then we apply the L’ Hopital’s rule which means that we will differentiate the numerator and the denominator separately and then we put the value. We will also apply the logarithm formula to solve this.
Formula used:
$\log {a^n} = n\log a$
$\dfrac{d}{{dx}}\ln (x) = \dfrac{1}{x}$

Complete step by step answer:
Here we have $f(x) = \left\{ {\dfrac{{\log {{(1 + x)}^{1 + x}} - x}}{{{x^2}}}} \right\}$
By applying the logarithm formula in the numerator, we can write
$f(x) = \mathop {\lim }\limits_{x \to 0} \left\{ {\dfrac{{(1 + x)\log (1 + x) - x}}{{{x^2}}}} \right\}$
Now let us check that if we put the value of $x = 0$ in the expression,
$f(x) = \mathop {\lim }\limits_{x \to 0} \left\{ {\dfrac{{(1 + 0)\log (1 + 0) - 0}}{{{0^2}}}} \right\}$
It is in the form of $ \Rightarrow \dfrac{{1 - \log 1 - 0}}{0} = \dfrac{0}{0}$
So we will apply the L’ Hopital’s rule, by differentiating the numerator and denominator separately.
In the numerator, we will apply the product rule,
$\dfrac{d}{{dx}}f(1)f(2) = f'(1)f(2) + f(1)f'(2)$
By comparing here we have $f(1) = 1 + x$ and $f(2) = \log (1 + x)$
Now we know that derivative of $1 + x$ is
$1$
And we will apply the formula:
$\dfrac{d}{{dx}}\ln (x) = \dfrac{1}{x}$
By comparing this we have $(1 + x)$ in place of $x$ .
 So by applying this, we can say that the derivative of
 $\log (1 + x) = \dfrac{1}{{1 + x}}$
Similarly in the denominator, the derivative of ${x^2}$ is
$2x$
By putting the values in the expression we have:
$\mathop {\lim }\limits_{x \to 0} \dfrac{{1 \times \log (1 + x) + \left( {1 + x} \right) \times \dfrac{1}{{(1 + x)}} - 1}}{{2x}}$
On simplifying it gives us
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\log (1 + x) + 1 - 1}}{{2x}}$
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\log (1 + x)}}{{2x}}$

Now we will again put the value $x = 0$ , it gives us
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\log (1 + 0)}}{{20}} = \dfrac{0}{0}$
So we can again apply the L’ Hopital’s rule:
Again the derivative of numerator is
 $\dfrac{1}{{1 + x}}$
And the derivative of numerator $2x$ , we will apply the power rule i.e.
$2 \times 1 \times {x^{1 - 1}}$
It gives the value
 $ \Rightarrow 2 \times {x^0} = 2$
Now we will put the values in the equation, so we have
 $\mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{1}{{1 + x}}}}{2}$
On simplifying we have:
$\mathop {\lim }\limits_{x \to 0} \dfrac{1}{{2(x + 1)}}$
We will put the value that $x \to 0$ , so it gives:
$ \Rightarrow \dfrac{1}{{2(1 + 0)}} = \dfrac{1}{2}$
Hence the required answer is $\dfrac{1}{2}$

Note:
We should note that we can use L’Hopital’s rule only if we have both numerator and denominator as zero and we can use this rule more than once if we are getting the same form of expression. There is no boundation like we can’t use it twice in solution.