Let $$f:{\bf{R}} \to {\bf{R}}$$ be a positive increasing function with $$\mathop {\lim }\limits_{x \to \infty } \frac{{f\left( {3x} \right)}}{{f\left( x \right)}} = 1.$$ Then $$\mathop {\lim }\limits_{x \to \infty } \frac{{f\left( {2x} \right)}}{{f\left( x \right)}} = ?$$
A.
$$\frac{2}{3}$$
B.
$$\frac{3}{2}$$
C.
$$3$$
D.
$$1$$
Answer :
$$1$$
Solution :
$$f\left( x \right)$$ is a positive increasing function.
$$\eqalign{
& \therefore 0 < f\left( x \right) < f\left( {2x} \right) < f\left( {3x} \right) \cr
& \Rightarrow 0 < 1 < \frac{{f\left( {2x} \right)}}{{f\left( x \right)}} < \frac{{f\left( {3x} \right)}}{{f\left( x \right)}} \cr
& \Rightarrow \mathop {\lim }\limits_{x \to \infty } \,1 \leqslant \mathop {\lim }\limits_{x \to \infty } \frac{{f\left( {2x} \right)}}{{f\left( x \right)}} \leqslant \mathop {\lim }\limits_{x \to \infty } \frac{{f\left( {3x} \right)}}{{f\left( x \right)}} \cr} $$
By Sandwich Theorem.
$$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } \frac{{f\left( {2x} \right)}}{{f\left( x \right)}} = 1$$
Releted MCQ Question on Calculus >> Limits
Releted Question 1
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