JEE MAIN - Mathematics (2010 - No. 2)
Let $$f:R \to R$$ be a positive increasing function with
$$\mathop {\lim }\limits_{x \to \infty } {{f(3x)} \over {f(x)}} = 1$$. Then $$\mathop {\lim }\limits_{x \to \infty } {{f(2x)} \over {f(x)}} = $$
$$\mathop {\lim }\limits_{x \to \infty } {{f(3x)} \over {f(x)}} = 1$$. Then $$\mathop {\lim }\limits_{x \to \infty } {{f(2x)} \over {f(x)}} = $$
$${2 \over 3}$$
$${3 \over 2}$$
3
1
Explanation
$$f(x)$$ is a positive increasing function.
$$\therefore$$ $$0 < f\left( x \right) < f\left( {2x} \right) < f\left( {3x} \right)$$
$$ \Rightarrow 0 < 1 < {{f\left( {2x} \right)} \over {f\left( x \right)}} < {{f\left( {3x} \right)} \over {f\left( x \right)}}$$
$$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } 1 \le \mathop {\lim }\limits_{x \to \infty } {{f\left( {2x} \right)} \over {f\left( x \right)}} \le \mathop {\lim }\limits_{x \to \infty } {{f\left( {3x} \right)} \over {f\left( x \right)}}$$
By Sandwich Theorem.
$$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {{f\left( {2x} \right)} \over {f\left( x \right)}} = 1$$
$$\therefore$$ $$0 < f\left( x \right) < f\left( {2x} \right) < f\left( {3x} \right)$$
$$ \Rightarrow 0 < 1 < {{f\left( {2x} \right)} \over {f\left( x \right)}} < {{f\left( {3x} \right)} \over {f\left( x \right)}}$$
$$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } 1 \le \mathop {\lim }\limits_{x \to \infty } {{f\left( {2x} \right)} \over {f\left( x \right)}} \le \mathop {\lim }\limits_{x \to \infty } {{f\left( {3x} \right)} \over {f\left( x \right)}}$$
By Sandwich Theorem.
$$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {{f\left( {2x} \right)} \over {f\left( x \right)}} = 1$$
Comments (0)
