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  • Evaluate the limit  <img alt="\lim _{x \rightarrow \frac{\pi}{2}} f\left(\frac{g(x)}{h(x)}\right)" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Clim%20_%7Bx%20%5Crightarrow%20%5Cfrac

Evaluate the limit  \lim _{x \rightarrow \frac{\pi}{2}} f\left(\frac{g(x)}{h(x)}\right) where h\left ( x \right )= \left ( \sin x \right )\left ( x^{2}-2x+4 \right ),g\left ( x \right )= \frac{x^{2}}{4}.

Option: 1

f\left(\frac{\frac{\pi^2}{16}}{\frac{\pi^2}{4}-\pi+4}\right)


Option: 2

f\left ( \frac{\frac{\pi^{2}}{16}}{\pi^{2}-\pi+4} \right )


Option: 3

f\left ( \frac{\pi^{2}}{\frac{\pi^{2}}{4}-\pi+4} \right )


Option: 4

f\left ( \frac{\frac{\pi^{2}}{4}}{\frac{\pi^{2}}{4}-\pi+4} \right )


Answers (1)

best_answer

The limit stated here is \lim _{x \rightarrow \frac{\pi}{2}} f\left(\frac{g(x)}{h(x)}\right) \quad \dots\left ( i \right )

where the following data is provided 

h\left ( x \right )= \left ( \sin x \right )\left ( x^{2}-2x+4 \right )\quad \dots\left ( ii \right )
g\left ( x \right )=\frac{x^{2}}{4}\quad \dots \left ( iii \right )

Note the following essential points.

  • The “Composition law of limit” states that the limit \lim_{x\rightarrow a}f\left ( g\left ( x \right ) \right )= f\left ( \lim_{x\rightarrow a}g\left ( x \right ) \right )= f\left ( A \right )  hold good, if and only if f\left ( x \right )   is continuous at g\left ( x \right )= A.
  • The “Quotient law for limits” states that \lim_{x\rightarrow a}\frac{f\left ( x \right )}{g\left ( x \right )}= \frac{\lim_{x\rightarrow a}f\left ( x \right )}{\lim_{x\rightarrow a}g\left ( x \right )}= \frac{A}{B}\, for\, B\neq 0
  • The function \sin \left ( x \right ) is continuous for \forall x \in R.


Use the equations (ii) and (iii), and apply the above laws of limit to rewrite the equation (i) in the following way.

\lim _{x \rightarrow \frac{\pi}{2}} f\left(\frac{g(x)}{h(x)}\right)
=f\left(\lim _{x \rightarrow \frac{\pi}{2}} \frac{g(x)}{h(x)}\right)
=f\left(\frac{\lim _{x \rightarrow \frac{\pi}{2}} g(x)}{\lim _{x \rightarrow \frac{\pi}{2}} h(x)}\right)
=f\left(\frac{\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{x^2}{4}\right)}{\lim _{x \rightarrow \frac{\pi}{2}}(\sin x)\left(x^2-2 x+4\right)}\right)
=f\left(\frac{\frac{\left(\frac{\pi}{2}\right)^2}{4}}{\left(\sin \frac{\pi}{2}\right)\left(\left(\frac{\pi}{2}\right)^2-2 \times \frac{\pi}{2}+4\right)}\right)
=f\left(\frac{\frac{\pi^2}{16}}{1 \times\left(\frac{\pi^2}{4}-\pi+4\right)}\right)
=f\left(\frac{\frac{\pi^2}{16}}{\frac{\pi^2}{4}-\pi+4}\right)

Posted by

Ritika Kankaria

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