Limit Continuity and Differentiability 1 Question 8
9. $\lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^{2} x\right)}{x^{2}}$ is equal to
(2014 Main)
(a) $\frac{\pi}{2}$
(b) 1
(c) $-\pi$
(d) $\pi$
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Answer:
Correct Answer: 9. (d)
Solution:
- PLAN To make the quadratic into simple form we should eliminate radical sign.
Description of Situation As for given equation, when $a \rightarrow 0$ the equation reduces to identity in $x$.
i.e. $a x^{2}+b x+c=0, \forall x \in R$ or $a=b=c \rightarrow 0$
Thus, first we should make above equation independent from coefficients as 0 .
Let $a+1=t^{6}$. Thus, when $a \rightarrow 0, t \rightarrow 1$.
$$ \begin{array}{rlrl} \therefore & \left(t^{2}-1\right) x^{2}+\left(t^{3}-1\right) x+(t-1) & =0 \\ \Rightarrow \quad(t-1){(t+1) x^{2}+\left(t^{2}+t+1\right) x+1 } & =0, \text { as } t \rightarrow 1 \\ 2 x^{2}+3 x+1 & =0 \\ \Rightarrow & & 2 x^{2}+2 x+x+1 & =0 \\ \Rightarrow & & (2 x+1)(x+1) & =0 \\ & \text { Thus, } & x & =-1,-1 / 2 \end{array} $$
or $\quad \lim _{a \rightarrow 0^{+}} \alpha(a)=-1 / 2$
and $\quad \lim _{a \rightarrow 0^{+}} \beta(a)=-1$
