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Evaluate the limit \lim _{x \rightarrow-2} f(g(x) h(x)) where h(x)=\frac{x^2-2 x+4}{x^2+3 x+5}, g(x)=\frac{x^3}{4}.

Option: 1

f(-18)


Option: 2

f(-8)


Option: 3

f(-4)


Option: 4

Cannot be determined


Answers (1)

best_answer

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

where the following data is provided
h(x)=\frac{x^2-2 x+4}{x^2+3 x+5} \quad \dots\left (i i \right )
g(x)=\frac{x^3}{4} \quad \dots\left (i ii \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-2} f(g(x) h(x))
=f\left(\lim _{x \rightarrow-2}(g(x) h(x))\right)
=f\left(\lim _{x \rightarrow-2} g(x) \times \lim _{x \rightarrow-2} h(x)\right)
=f\left(\lim _{x \rightarrow-2}\left(\frac{x^3}{4}\right) \times \lim _{x \rightarrow-2}\left(\frac{x^2-2 x+4}{x^2+3 x+5}\right)\right)
=f\left(\left(\frac{(-2)^3}{4}\right) \times\left(\frac{(-2)^2-2 \times(-2)+4}{(-2)^2+3 \times(-2)+5}\right)\right)
=f\left(\left(\frac{-8}{4}\right) \times\left(\frac{4+4+4}{4-6+5}\right)\right)
=f\left(-2 \times \frac{12}{3}\right)
=f(-8)


 

Posted by

himanshu.meshram

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