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

Option: 1

f\left ( \frac{\pi^{2}}{16}+1 \right )


Option: 2

f\left ( \frac{\pi^{2}}{4}+1 \right )


Option: 3

f\left ( \frac{\pi^{2}}{8}+1 \right )


Option: 4

f\left ( \frac{\pi^{2}}{4}+1 \right )


Answers (1)

best_answer

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

where the following data is provided

h\left ( x \right )= \sin x \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 “Sum law for limits” states that \lim_{x\rightarrow a}f\left ( x \right )+\lim_{x\rightarrow a}g\left ( x \right )= \lim_{x\rightarrow a}\left [ f\left ( x \right )+g\left ( x \right )\right ]
  • The function \sin \left ( x \right )  is continuous for \forall x \in R.

Use the equations (ii) and (iii), and apply the “Composition law of limit” to rewrite the equation (i) in the following way.

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

Posted by

Ajit Kumar Dubey

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