Question 298 of 401MediumMCQ
Let \[f:R \to R\] be a positive increasing function with
\[\mathop {\lim }\limits_{x \to \infty } {{f(3x)} \over {f(x)}} = 1\]. Then \[\mathop {\lim }\limits_{x \to \infty } {{f(2x)} \over {f(x)}} = \]
Solutions · 1
Official solution
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.
\[ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {{f\left( {2x} \right)} \over {f\left( x \right)}} = 1\] Answer the question to unlock