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

Question

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

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

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

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

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.

Answer 1

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.

Answer 2

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

Answer 3

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

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

Answer 4

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