Question

If it is given that $f:R\to R$ is a differentiable function such that $f\left( 0 \right)=0,f\left( \dfrac{\pi }{2} \right)=3\,and\,f'\left( 0 \right)=1$. If $g\left( x \right)=\int\limits_{x}^{\dfrac{\pi }{2}}{[f'\left( t \right)\operatorname{cosect}-\cot t.\operatorname{cosect}}.f\left( t \right)]dt\,\,for\,\,x\in \left( 0,\dfrac{\pi }{2} \right]$ then what will be the value of$\underset{x\to 0}{\mathop{\lim }}\,g\left( x \right)$?

Answer 1

Hint: To solve this problem we will apply the reverse of the product rule of the differentiation i.e.
the derivative of fg = f g’ + f’ g so also, f g’ + f’ g = (fg)’. After that we will apply the L’hospital’s rule,
It says that the limit when we divide one function by another is the same after we take the derivative of the each function, if the limit is of the form $\dfrac{0}{0}\,or\,\dfrac{\infty }{\infty }$.

Complete step-by-step answer:
We are given that,
$f\left( 0 \right)=0,f\left( \dfrac{\pi }{2} \right)=3\,and\,f'\left( 0 \right)=1$, and
$g\left( x \right)=\int\limits_{x}^{\dfrac{\pi }{2}}{[f'\left( t \right)\operatorname{cosect}-\cot t.\operatorname{cosect}}.f\left( t \right)]dt\,\,for\,\,x\in \left( 0,\dfrac{\pi }{2} \right]$.
Now if we observe $f'\left( t \right)\operatorname{cosect}-\cot t.\operatorname{cosect}.f\left( t \right)$ it is the derivative of $f\left( t \right).\operatorname{cosect}$, i.e. we get
$\dfrac{d\left( f\left( t \right).\operatorname{cosect} \right)}{dt}=f'\left( t \right)\operatorname{cosect}-\cot t.\operatorname{cosect}.f\left( t \right)$
Above shown equation we get is kind of reverse of the product rule of differentiation.
Now putting $f'\left( t \right)\operatorname{cosect}-\cot t.\operatorname{cosect}.f\left( t \right)=\dfrac{d\left( f\left( t \right).\operatorname{cosect} \right)}{dt}$ in the function g(x), we get
$g\left( x \right)=\int\limits_{x}^{\dfrac{\pi }{2}}{\dfrac{d\left( f\left( t \right).\operatorname{cosect} \right)}{dt}dt}$
\[\begin{align}
  & g\left( x \right)=\left[ \left( f\left( t \right).\operatorname{cosect} \right) \right]_{x}^{\dfrac{\pi }{2}} \\
 & g\left( x \right)=f\left( \dfrac{\pi }{2} \right).\operatorname{cosec}\dfrac{\pi }{2}-f\left( x \right).\operatorname{cosec}x \\
\end{align}\]
Now we will find the limit $\underset{x\to 0}{\mathop{\lim }}\,g\left( x \right)$, as
$\underset{x\to 0}{\mathop{\lim }}\,g\left( x \right)=\underset{x\to 0}{\mathop{\lim }}\,\left( f\left( \dfrac{\pi }{2} \right).\operatorname{cosec}\dfrac{\pi }{2}-f\left( x \right).\operatorname{cosec}x \right)$
$\underset{x\to 0}{\mathop{\lim }}\,g\left( x \right)=f\left( \dfrac{\pi }{2} \right).\operatorname{cosec}\dfrac{\pi }{2}-\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right).\operatorname{cosec}x$
As we know $\operatorname{cosec}x=\dfrac{1}{\sin x}$,
$\underset{x\to 0}{\mathop{\lim }}\,g\left( x \right)=f\left( \dfrac{\pi }{2} \right).\operatorname{cosec}\dfrac{\pi }{2}-\underset{x\to 0}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{\sin x}$
Now limit $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{\sin x}$ is of the form $\dfrac{0}{0}$ because $f\left( 0 \right)=0$ is given, hence we can apply L’hospital’s rule here,
L’hospital says that if we have an indeterminate form $\dfrac{0}{0}\,or\,\dfrac{\infty }{\infty }$, all we need to do is differentiate the numerator and differentiate the denominator and then take the limit. So applying the L’hospital’s rule, we get
$\underset{x\to 0}{\mathop{\lim }}\,g\left( x \right)=f\left( \dfrac{\pi }{2} \right).\operatorname{cosec}\dfrac{\pi }{2}-\underset{x\to 0}{\mathop{\lim }}\,\dfrac{f'\left( x \right)}{\cos x}$
Now, putting the values $f\left( \dfrac{\pi }{2} \right)=3\,,f'\left( 0 \right)=1,\operatorname{cosec}\dfrac{\pi }{2}=1\,and\,\cos 0=1$, we get
$\underset{x\to 0}{\mathop{\lim }}\,g\left( x \right)=3-1=2$
$\underset{x\to 0}{\mathop{\lim }}\,g\left( x \right)=2$
Hence our answer for the value of the limit $\underset{x\to 0}{\mathop{\lim }}\,g\left( x \right)$ is 2.

Note: Students may make mistake while applying the L’hospital’s rule and may not check the form of the limit i.e. we can apply L’hospital’s rule only if the limit is of the form $\dfrac{0}{0}\,or\,\dfrac{\infty }{\infty }$ else we cannot apply this rule. And you may also do not change \[\operatorname{cosec}t\] to $\dfrac{1}{\sin t}$ in the function g(x) and remain stuck in order to solve the limit so usually try to convert secant, cosecant to sine and cosine.