Question

If f(x) is continuous and $f(\dfrac{9}{2}) = \dfrac{2}{9}$, then $\mathop {\lim }\limits_{x \to 0} f(\dfrac{{1 - \cos 3x}}{{{x^2}}})$ is equal to:-
A. $\dfrac{9}{2}$
B. $\dfrac{2}{9}$
C. 0
D. $\dfrac{8}{9}$

Answer 1

This question can be done by concept of continuous functions. Brief explanation is mentioned below: -
Continuous functions: - A function is said to be continuous at point $x = c$, if function exists at that point and is given by: -
$\mathop {\lim }\limits_{x \to c} f(x)$ Exists
$\mathop {\lim }\limits_{x \to c} f(x) = f(c)$

Complete step-by-step answer:
First of all we will check the form of limit by putting x=o in the function.
$\mathop {\lim }\limits_{x \to 0} f(\dfrac{{1 - \cos 3x}}{{{x^2}}})$
$\Rightarrow f(\dfrac{{1 - \cos 3(0)}}{0})$
$\Rightarrow f(\dfrac{{1 - 1}}{0})$ (Because $\cos 0 = 1$)
$\Rightarrow f(\dfrac{0}{0})$As function is 0/0 form now we will apply L Hospital’s rule in this numerator and denominator are individually differentiated with respect to ‘x’ till 0/0 form is not observed and then limit’s value is placed.
$\Rightarrow \mathop {\lim }\limits_{x \to 0} f\dfrac{{\dfrac{{d(1 - \cos 3x)}}{{dx}}}}{{\dfrac{{d{x^2}}}{{dx}}}}$
$\Rightarrow \mathop {\lim }\limits_{x \to 0} f[\dfrac{{3(\sin 3x)}}{{2x}}]$
Now again we will check 0/0 form and if it still comes then we will differentiate again.
$\Rightarrow \mathop {\lim }\limits_{x \to 0} f[\dfrac{{3(\sin 3(0))}}{{2(0)}}] = \dfrac{0}{0}$
Now we will derive one more time to eliminate 0/0 form
$\Rightarrow \mathop {\lim }\limits_{x \to 0} f[\dfrac{{\dfrac{{{d^2}3(\sin 3x)}}{{d{x^2}}}}}{{\dfrac{{{d^2}2x}}{{d{x^2}}}}}]$

$\Rightarrow \mathop {\lim }\limits_{x \to 0} f[\dfrac{{9\cos 3x}}{2}]$
Now we will put a limit inside the function.
$\Rightarrow f[\dfrac{{9\cos 3(0)}}{2}] = f(\dfrac{9}{2})$
Now given in question that$f(\dfrac{9}{2}) = \dfrac{2}{9}$

So, the correct answer is “Option B”.

Note: Students may likely make mistakes while putting limits directly into the function without checking its form so firstly check the function's form then apply L'Hospital's rule. Also students can easily do differentiation by applying chain rule. There is one example to show how to apply chain rule.
For example: - $f(x) = 4\sin 3x$ this is the function now we will apply chain rule. Here firstly write the constant part as it is then go to another term sin3x, as we all know differentiation of sinx is cosx so differentiation of sin3x will be cos3x then go more inside there is 3x whose differentiation will be 3.
Final answer will be $f'(x) = 4(3)\cos 3x = 12\cos 3x$