# What's the value of k given f(x) is continuous at x = 0 without using L hospital rule?

Just recently, i was asked to solve this question in my exam. This is the following question.

Given $f(x) = \left\{ \begin{array}{ll} \frac{\sin(\pi (\cos(x))^2)}{x^2}, & \text{ if } x \not= 0 \\ k, & \text{ if } x = 0 \end{array} \right.$

is continuous at x = 0. Find the value of k.

Since, we know that this function is continuus at x = 0. We can imply that, $$\;\lim_{h\to0}\frac{\sin (\pi (\cos(h))^2)}{h^2}=k$$

I computed the limit using L'hospital rule and it was $\pi$, but i was wondering how to compute it without the use of L'hospital rule.

• well, you can try Taylor's expansion of the numerator. – OnoL Feb 9 '18 at 17:45
• @OnoL I forgot to mention, i can't use it since my teacher told me to solve it using what was covered in syllabus. And Apparently, there are no Taylor expansions and L'hospital rule in my syllabus :( – Utkarsh Dixit Feb 9 '18 at 17:47
• Correct me if I'm wrong, but if the function is continuous at $x=0$ then can't you set the previous equation to 0 and solve from there? – Gil Keidar Feb 9 '18 at 17:47
• @GilKeidar He needs to find $k$ so that it is continuous. It is when $k=\pi$. – Atmos Feb 9 '18 at 17:48
• @OnoL For analytic functions, L'H is a trivial consequence of the existence of the Taylor series. I would be very hesitant to use Taylor series if L'H is not allowed. – N. S. Feb 9 '18 at 17:58

Using $\sin(\pi-x)=\sin(x)$ we get $$\;\lim_{h\to0}\frac{\sin (\pi (\cos(h))^2)}{h^2}=\lim_{h\to0}\frac{\sin (\pi- \pi (\cos(h))^2)}{h^2}\\ =\lim_{h\to0}\frac{\sin (\pi-\pi (\cos(h))^2)}{\pi- \pi (\cos(h))^2}\cdot \pi \cdot\frac{1- \cos(h)^2}{h^2}=\pi\lim_{h\to0}\frac{\sin (\pi-\pi (\cos(h))^2)}{\pi- \pi (\cos(h))^2}\cdot\frac{\sin(h)^2}{h^2}$$

Now both limits are 1 by the fundamental trig limit.

We can use Taylor expansions $$\frac{\sin\left(\pi\left(\cos(h)\right)^2\right)}{h^2}=\frac{\sin\left(\pi\left(1-h^2/2+o\left(h^2\right)\right)^2\right)}{h^2}=\frac{\sin\left(\pi-\pi h^2 +o\left(h^2\right)\right)}{h^2}\underset{(0)}{\sim}\frac{\sin\left(\pi h^2\right)}{h^2}$$ Hence

$$\frac{\sin\left(\pi\left(\cos(h)\right)^2\right)}{h^2}\underset{(0)}{\sim}\frac{\pi h^2}{h^2}=\pi$$

• Thanks @Atmos, but i can't use Taylor expansion but, it's a neat answer. +1 – Utkarsh Dixit Feb 9 '18 at 17:52

$\frac {sin(\pi(cosx)^2)}{x^2}=\frac {sin(\pi-\pi(cosx)^2)}{x^2}=\frac {sin(\pi(sinx)^2)}{x^2}=\frac {sin(\pi(sinx)^2)}{\pi(sinx)^2}\frac{\pi(sinx)^2}{x^2}$ Now as $x\rightarrow 0$ the limit is $\pi$.So $k= \pi.$

$\dfrac{\sin(π(1-\sin^2h))}{h^2} =$

$\dfrac{-\cos(π)\sin(π\sin^2h)}{h^2}=$

$\dfrac{\sin(π\sin^2h)}{π\sin^2h} \dfrac{π\sin^2h}{h^2}.$

Limit:

1)$\lim_{h \rightarrow 0} \dfrac{\sin(π\sin^2h)}{π\sin^2h}=1;$

2)$\lim_{h \rightarrow 0} \dfrac{π\sin^2h}{h^2}=π$