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

Hint:
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.