Question

If f(x)=0 is a quadratic equation such that $\rm{f(-π)=f(π)=0 \: and \: f( \frac{\pi}{2} ) = - \frac{3 {\pi}^{2} }{4} }$
the find the value of $\displaystyle\rm \lim_{x \to0} \frac{f(x)}{ \sin(sinx) }$ ​

Answer 1

Answer:

Step-by-step explanation:

We have,

f(x) is a quadratic equation, so,

$\sf{Let\,\,\,\,f(x)=a{x}^{2}+b{x}+c}$

And, we are given,

$\sf{f(-\pi)=f(\pi)=0\,\,\,\,\,\&\,\,\,\,\,f\left(\dfrac{\pi}{2}\right)=-\dfrac{3{\pi}^{2}}{4}}$

So,

$\sf{\dfrac{a{\pi}^{2}}{4}+\dfrac{b\pi}{2}+c=-\dfrac{3{\pi}^{2}}{4}}$

$\sf{\implies\,a{\pi}^{2}+2b\pi+4c=-3{\pi}^{2}}$

$\sf{\implies\,(a+3){\pi}^{2}+2b\pi+4c=0\,\,\,\,\,\,\,\,\,\,\,\,\,...(1)}$

$\sf{\implies\,a{\pi}^{2}-b{\pi}+c=0}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(2)$

$\sf{\implies\,a{\pi}^{2}+b{\pi}+c=0}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(3)$

on solving these equations, we get,

$\tt{a=1\,\,\,\,,\,\,\,\,b=0\,\,\,\,,\,\,\,\,c=-\pi^2}$

So,

$\boxed{\sf{f(x)={x}^{2}-\pi^2}}$

Now,

$\displaystyle\lim_{x\to0}\dfrac{f(x)}{\sin\left(\sin(x)\right)}$

$\displaystyle=\lim_{x\to0}\dfrac{\sin(x)}{\sin\left(\sin(x)\right)}\cdot\dfrac{f(x)}{\sin(x)}$

$\displaystyle=\lim_{\sin(x)\to0}\dfrac{\sin(x)}{\sin\left(\sin(x)\right)}\cdot\lim_{x\to0}\dfrac{x}{\sin(x)}\cdot\lim_{x\to0}\dfrac{f(x)}{x}$

$\displaystyle=\lim_{\sin(x)\to0}\dfrac{\sin(x)}{\sin\left(\sin(x)\right)}\cdot\lim_{x\to0}\dfrac{x}{\sin(x)}\cdot\lim_{x\to0}\dfrac{{x}^{2}-{\pi}^{2}}{x}$

on the last limit, apply l'hospital's rule,

$\displaystyle=\lim_{\sin(x)\to0}\dfrac{\sin(x)}{\sin\left(\sin(x)\right)}\cdot\lim_{x\to0}\dfrac{x}{\sin(x)}\cdot\lim_{x\to0}\dfrac{2x}{1}$

$\displaystyle=1\cdot1\cdot\lim_{x\to0}\dfrac{2x}{1}$

$=0$