First slide

Introduction to limits

Question

If f(x) = 0 is a quadratic equation such that $f (- π) = f (π) = 0 and f (\frac{π}{2}) = - \frac{3 π^{2}}{4},$ then $\lim_{x \to - π} \frac{f (x)}{\sin (sinx)}$ is equal to

Moderate

A

0
B

$π$
C

$2 π$
D

none of these

Solution

$\begin{array}{l} Given f (x) = x^{2} - π^{2} \\ \lim_{x \to - π} \frac{x^{2} - π^{2}}{\sin (sinx)} = \lim_{h \to 0} \frac{{(- π + h)}^{2} - π^{2}}{\sin (\sin (- π + h))} \\ = \lim_{h \to 0} \frac{- 2 hπ + h^{2}}{- \sin (\sin h)} \\ = \lim_{h \to 0} \frac{h - 2 π}{\frac{- \sin (\sin h)}{\sin h} \times \frac{\sin h}{h}} = 2 π \end{array}$

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