Question

A function $f(x)$is defined as below $f(x)$ $= \dfrac{{\cos (\sin x) - \cos x}}{{{x^2}}},\,\,x \ne 0$ and $f(0)$ $= a$ $f(x)$is continuous at $x = 0$ if $a$ equals.

Answer 1

In the Mathematics, a continuous function is a function that does not have any abrupt changes in value known as discontinuities. Efficiently small changes in the input of a continuous function result in orbit rarity small changes in its output. If not continuous a function is said to be discontinuous.
L'Hospital's Rule tells us that if we have an indeterminate form $\dfrac{0}{0}$or $\dfrac{\infty }{\infty }$ all we need to do is differentiate the numerator and differentiate the denominator and then take the limit.

Complete step by step solution:
Given $f(x)$ is continuous at $n = 0$ then $f(x)$ must be equal to the limit value at $x \to 0$.
$f(x) = \lim \,f(x)$.
$x \to 0$
Since $f(0) = a$
$\therefore \,a = \lim \,\dfrac{{\cos (\sin x) - \cos x}}{{\mathop x\nolimits^2 }}\,$
$x \to 0$
Now, we will put $x = 0$ in$f(x)$.
$\to a = \dfrac{0}{0}$ From
Now, we apply the L- Hospital Rule.
In L-Hospital we differentiate $f(x)$ i.e. Both Numerator and Denominator.
Separately,
$\therefore$ After Applying L-hospital Rule
$\lim \dfrac{{ - \sin (\sin x).\cos x + \sin x}}{{2x}}$
Now, again put $x \to 0$.
Again $\dfrac{{0 + 0}}{0}$ from is Available.
So, we will now again Apply L-Hospital Rule.
$\to \lim = \dfrac{{\cos \,(\sin x).\cos x + \sin x(\sin x)\sin x + \cos }}{2}$
Here, differentiate done by Product rule
In product rule. Eg. If there ‘a, b’ in product then differentiate of
$\dfrac{d}{{dx}}(a.b) = a{b^1} + b{a^1}$
${a^1},{b^1}$ are the differentiated part of both a and b Respectively.
$\Rightarrow \,\dfrac{{ - \cos (\sin \theta ).{{\cos }^2}(0) + \sin \,(\sin \theta ).\sin \theta + \cos \theta }}{2}$
$\Rightarrow \,\dfrac{{ - \cos (0){{.1}^2} + \sin \,(0).0 + 1}}{2}$
$\Rightarrow \,\dfrac{{ - 1 + 0 + 1}}{2} = \dfrac{0}{2} = 0$.
$\therefore \,a = 0$.

Note: Continuity has a limited No. of solution is to solve because contimity is a very described way of solution. If continuity is proved then only L-Hospital Rule is proved then only L-Hospital Rule is the only way to prove the solution.