Question

For what values of ‘k’ is the function f(x) = \left\\{ \begin{aligned} & \dfrac{\sin 5x}{3x}+\cos x,\,\,if\,x\ne 0 \\\ & k,\,\,if\,x=0 \\\ \end{aligned} \right. continuous at x = 0?

Answer 1

To solve this problem we need to know that for a function f(x) to be continuous at x = ‘a’ then for that limit of the function should exist at that point i.e. left hand limit and right hand limit both should be equal at x = ‘a’ and also the obtained limit should be equal to value of that function at the point ‘a’. So to solve this we will first find the limit of the function at x = 0 and then equate that obtained limit with the value of the function at x = 0. And from that we will get the correct answer.

Complete step-by-step answer:
To solve this we have to know that for a function to be continuous at a given point then its limit should exist at that point and that limit should be equal to the value of the function at that point.
So first we will calculate the limit of the function at the given point i.e. x = 0,
We are given that,
f(x) = \left\\{ \begin{aligned} & \dfrac{\sin 5x}{3x}+\cos x,\,\,if\,x\ne 0 \\\ & k,\,\,if\,x=0 \\\ \end{aligned} \right.
we will find the limit at x = 0 as following,
since we are given that f(x) is continuous at x = 0, so
$\displaystyle \lim_{x \to 0}\left( \dfrac{\sin 5x}{3x}+\cos x \right)$
Multiplying the term $\dfrac{\sin 5x}{3x}$ with $\dfrac{5x}{5x}$ we get,
$\displaystyle \lim_{x \to 0}\left( \dfrac{\sin 5x}{5x}.\dfrac{5x}{3x}+\cos x \right)$
$\displaystyle \lim_{x \to 0}\left( \dfrac{\sin 5x}{5x} \right)\dfrac{5}{3}+\left( \displaystyle \lim_{x \to 0}\cos x \right)$
We know that $\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}=1$, so applying this we get,
$\dfrac{5}{3}+\left( \displaystyle \lim_{x \to 0}\cos x \right)$
$\dfrac{5}{3}+1=\dfrac{8}{3}$
So the limit of the function at x = 0 is $\dfrac{8}{3}$.
Since it is given that f(x) is continuous at x = 0, so its limit at x = 0 will be equal to its value at x = 0,
And its value at x = 0 is equal to k as it is already given,
So we get
$k=\dfrac{8}{3}$
Hence we get our answer as $\dfrac{8}{3}$.

Note: You should note and remember that if the function is continuous at a point ‘a’ then it its limit must exist at that point whereas converse is not true i.e. if the limit exist at point ‘a’ it may or maynot be continuous at that point. Also in the above question remember to equate the obtained limit with the value of that function some students do not equate and directly write the answer but it is wrong you have to equate it with the value to get the answer.