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Evaluate the limit \lim_{x\rightarrow \frac{x}{2}}f\left ( g\left ( h\left ( x \right ) \right ) \right ) where h\left ( x \right )= \sin x,g\left ( x \right )= x^{3}.

Option: 1

f\left ( 1 \right )


Option: 2

g\left ( \frac{\pi}{2} \right )


Option: 3

f\left ( \frac{\pi}{2} \right )


Option: 4

h\left ( 1 \right )


Answers (1)

best_answer

The limit stated here is \lim_{x\rightarrow \frac{x}{2}}f\left ( g\left ( h\left ( x \right ) \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 )= x^{3} \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 functions \sin \left ( x \right ) and \cos \left ( x \right ) are both 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(h(x)))
=f\left(\lim _{x \rightarrow \frac{\pi}{2}} g(h(x))\right)
=f\left(g\left(\lim _{x \rightarrow \frac{\pi}{2}} h(x)\right)\right)
=f\left(g\left(\lim _{x \rightarrow \frac{\pi}{2}} \sin x\right)\right)
=f\left(g\left(\sin \left(\frac{\pi}{2}\right)\right)\right)
=f(g(1))
=f\left(1^3\right)
=f(1)
 





 

Posted by

vishal kumar

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