Question

Find the value of a for which the function f defined by $f\left( x \right)=\left\{ \begin{matrix}
   a\sin \left( \dfrac{\pi }{2}\left( x+1 \right) \right),x\le 0 \\
   \dfrac{\tan x-\sin x}{{{x}^{3}}},x>0 \\
\end{matrix} \right.$ is continuous at x = 0.
[a] $\dfrac{1}{2}$
[b] $\dfrac{-1}{2}$
[c] 2
[d] -2

Answer 1

Hint: Use the fact that if f(x) is continuous at $x={{x}_{0}}$, then $\displaystyle \lim_{x \to x_{0}^{+}}f\left( x \right)=\displaystyle \lim_{x \to x_{0}^{+}}f\left( x \right)=f\left( 0 \right)$. Hence prove that $a=\displaystyle \lim_{x \to 0}\dfrac{\tan x-\sin x}{{{x}^{3}}}$. Use the fact that $\tan x=\dfrac{\sin x}{\cos x}$ and hence prove that \[a=\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}\times \dfrac{1-\cos x}{{{x}^{2}}}\times \dfrac{1}{\cos x}\]. Use the fact that if $\displaystyle \lim_{x \to {{x}_{0}}}f\left( x \right)=l$ and $\displaystyle \lim_{x \to {{x}_{0}}}g\left( x \right)=m$, then $\displaystyle \lim_{x \to {{x}_{0}}}f\left( x \right)g\left( x \right)=ml$. Hence determine the value of $\displaystyle \lim_{x \to 0}\dfrac{\tan x-\sin x}{{{x}^{3}}}$. Use the results $\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}=1$ and $\displaystyle \lim_{x \to 0}\dfrac{1-\cos x}{{{x}^{2}}}=\dfrac{1}{2}$. Hence find the value of a.

Complete step by step answer:
We have
$f\left( x \right)=\left\{ \begin{matrix}
   a\sin \left( \dfrac{\pi }{2}\left( x+1 \right) \right),x\le 0 \\
   \dfrac{\tan x-\sin x}{{{x}^{3}}},x>0 \\
\end{matrix} \right.$ is continuous at x = 0.
We know that if f(x) is continuous at $x={{x}_{0}}$, then $\displaystyle \lim_{x \to x_{0}^{+}}f\left( x \right)=\displaystyle \lim_{x \to x_{0}^{+}}f\left( x \right)=f\left( 0 \right)$.
Hence, we have
$\displaystyle \lim_{x \to {{0}^{+}}}f\left( x \right)=f\left( 0 \right)=a\sin \left( \dfrac{\pi }{2} \right)=a$
Hence, we have
$a=\displaystyle \lim_{x \to {{0}^{+}}}\dfrac{\tan x-\sin x}{{{x}^{3}}}$
We know that $\displaystyle \lim_{x \to {{a}^{+}}}f\left( x \right)=\displaystyle \lim_{h\to 0}f\left( a+h \right)$
Hence, we have
$a=\displaystyle \lim_{h\to 0}\dfrac{\tan \left( h \right)-\sin \left( h \right)}{{{h}^{3}}}$
We know that $\tan \left( h \right)=\dfrac{\sin \left( h \right)}{\cos \left( h \right)}$
Hence, we have
$a=\displaystyle \lim_{h\to 0}\dfrac{\dfrac{\sin \left( h \right)}{\cos \left( h \right)}-\sin \left( h \right)}{{{h}^{3}}}$
Multiplying numerator and denominator by $\cos \left( h \right)$, we get
\[a=\displaystyle \lim_{h\to 0}\dfrac{\sin \left( h \right)-\cos \left( h \right)\sin \left( h \right)}{{{h}^{3}}\cos \left( h \right)}\]
Taking $\sin \left( h \right)$ common from the terms in numerator, we get
$a=\displaystyle \lim_{h\to 0}\dfrac{\sin \left( h \right)\left( 1-\cos \left( h \right) \right)}{{{h}^{3}}\cos \left( h \right)}$
Writing ${{h}^{3}}$ as ${{h}^{2}}h$, we get
$a=\displaystyle \lim_{h\to 0}\dfrac{\sin \left( h \right)\left( 1-\cos \left( h \right) \right)}{h\times {{h}^{2}}\cos \left( h \right)}$
We know that $\dfrac{ab}{cd}=\dfrac{a}{c}\times \dfrac{b}{d}$
Hence, we have
$a=\displaystyle \lim_{h\to 0}\dfrac{\sin \left( h \right)}{h}\times \dfrac{1-\cos \left( h \right)}{{{h}^{2}}}\times \dfrac{1}{\cos \left( h \right)}$
We know that $\displaystyle \lim_{h\to 0}\dfrac{\sin \left( h \right)}{h}=1$
We know that $1-\cos \left( h \right)=2{{\sin }^{2}}\left( \dfrac{h}{2} \right)$
Hence, we have $\dfrac{1-\cos \left( h \right)}{{{h}^{2}}}=\dfrac{2{{\sin }^{2}}\left( \dfrac{h}{2} \right)}{{{h}^{2}}}=2{{\left( \dfrac{\sin \left( \dfrac{h}{2} \right)}{h} \right)}^{2}}$
Dividing numerator and denominator of the squared term by 2, we get
$\dfrac{1-\cos \left( h \right)}{{{h}^{2}}}=2{{\left( \dfrac{1}{2}\dfrac{\sin \left( \dfrac{h}{2} \right)}{\left( \dfrac{h}{2} \right)} \right)}^{2}}=\dfrac{1}{2}{{\left( \dfrac{\sin \left( \dfrac{h}{2} \right)}{\dfrac{h}{2}} \right)}^{2}}$
Hence, we have
\[\displaystyle \lim_{h\to 0}\dfrac{1-\cos \left( h \right)}{{{h}^{2}}}=\dfrac{1}{2}\left( \displaystyle \lim_{h\to 0}\dfrac{\sin \left( \dfrac{h}{2} \right)}{\dfrac{h}{2}} \right)=\dfrac{1}{2}\]
Also, we have $\displaystyle \lim_{h\to 0}\cos \left( h \right)=1\Rightarrow \displaystyle \lim_{h\to 0}\dfrac{1}{\cos \left( h \right)}=\dfrac{1}{1}=1$
We know that if $\displaystyle \lim_{x \to {{x}_{0}}}f\left( x \right)=l$ and $\displaystyle \lim_{x \to {{x}_{0}}}g\left( x \right)=m$, then $\displaystyle \lim_{x \to {{x}_{0}}}f\left( x \right)g\left( x \right)=ml$
Hence, we have
$a=\displaystyle \lim_{h\to 0}\dfrac{\sin \left( h \right)}{h}\times \displaystyle \lim_{h\to 0}\dfrac{1-\cos \left( h \right)}{{{h}^{2}}}\times \displaystyle \lim_{h\to 0}\dfrac{1}{\cos \left( h \right)}=\dfrac{1}{2}$
Hence the value of a is $\dfrac{1}{2}$
Hence option [a] is correct.

Note:
[1] Alternatively we can evaluate the limit using the series expansions of tanx and sinx.
We know that $\sin x=x-\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{5}}}{5!}+O\left( {{x}^{7}} \right)$ and $\tan x=x+\dfrac{{{x}^{3}}}{3}+\dfrac{{{x}^{5}}}{5}+O\left( {{x}^{7}} \right)$
Hence, we have
$\begin{align}
  & \tan x-\sin x=x+\dfrac{{{x}^{3}}}{3}+\dfrac{{{x}^{5}}}{5}-x+\dfrac{{{x}^{3}}}{6}-\dfrac{{{x}^{5}}}{120}+O\left( {{x}^{7}} \right) \\
 & =\dfrac{{{x}^{3}}}{2}+\dfrac{23{{x}^{5}}}{120}+O\left( {{x}^{7}} \right) \\
\end{align}$
Hence, we have
$\dfrac{\tan x-\sin x}{{{x}^{3}}}=\dfrac{\dfrac{{{x}^{3}}}{2}+\dfrac{23{{x}^{5}}}{120}+O\left( {{x}^{7}} \right)}{{{x}^{3}}}$
We know that $\dfrac{a+c}{b}=\dfrac{a}{b}+\dfrac{c}{b}$.
Using the above identity, we get
$\dfrac{\tan x-\sin x}{{{x}^{3}}}=\dfrac{{{x}^{3}}}{2{{x}^{3}}}+\dfrac{23{{x}^{5}}}{120{{x}^{3}}}+\dfrac{O\left( {{x}^{7}} \right)}{{{x}^{3}}}$
We know that $\dfrac{{{x}^{a}}}{{{x}^{b}}}={{x}^{a-b}}$
Hence, we have
$\dfrac{\tan x-\sin x}{{{x}^{3}}}=\dfrac{1}{2}+\dfrac{23}{120}{{x}^{2}}+O\left( {{x}^{4}} \right)$
Hence, we have
$\displaystyle \lim_{x \to 0}\dfrac{\tan x-\sin x}{{{x}^{3}}}=\dfrac{1}{2}+0+0=\dfrac{1}{2}$, which is the same as obtained above.
Hence $a=\dfrac{1}{2}$ and hence option [a] is correct.