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Evaluate the limit \lim _{x \rightarrow \frac{3 \pi}{2}} f(g(h(x))) where h(x)=\sin x, g(x)=10^3\left(x^3+\cos x\right)  [angles in degrees].

Option: 1

f(-1.16)


Option: 2

f(0)


Option: 3

f(2.16)


Option: 4

f(-0.16)


Answers (1)

best_answer

The limit stated here is \lim _{x \rightarrow \frac{3 \pi}{2}} f(g(h(x))) \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 )= 10^{3}\left ( x^{3}+\cos x \right ) \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 function \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{3 \pi}{2}} f(g(h(x)))
=f\left(\lim _{x \rightarrow \frac{3 \pi}{2}} g(h(x))\right)
=f\left(g\left(\lim _{x \rightarrow \frac{3 \pi}{2}} h(x)\right)\right)
=f\left(g\left(\lim _{x \rightarrow \frac{3 \pi}{2}} \sin x\right)\right)
=f\left(g\left(\sin \left(\frac{3 \pi}{2}\right)\right)\right)
=f(g(-1))
=f\left(10^3\left((-1)^3+\cos 1^0\right)\right)
=f\left(10^3(-1+0.99984)\right)
=f\left(10^3 \times(-0.00016)\right)
=f(-0.16)
 

Posted by

Pankaj

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