Let f: R^{+} rightarrow R be an increasing fu | vedclass

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(A) Given that $f$ is an increasing function and $f(x) > 0$ for all $x \in R^{+}$.We are given that $\lim _{x \rightarrow \infty} \frac{f(9 x)}{f(3 x)

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(A) Given that $f$ is an increasing function and $f(x) > 0$ for all $x \in R^{+}$.We are given that $\lim _{x \rightarrow \infty} \frac{f(9 x)}{f(3 x)}=1$.Since $f$ is an increasing function,for any $x > 0$,we have the inequality:$3x $\Rightarrow f(3x) Dividing the entire inequality by $f(3x)$ (which is positive),we get:$\frac{f(3x)}{f(3x)} $1 Now,taking the limit as $x \rightarrow \infty$ on all sides:$\lim _{x \rightarrow \infty} 1 \leq \lim _{x \rightarrow \infty} \frac{f(6x)}{f(3x)} \leq \lim _{x \rightarrow \infty} \frac{f(9x)}{f(3x)}$$1 \leq \lim _{x \rightarrow \infty} \frac{f(6x)}{f(3x)} \leq 1$By the Sandwich Theorem (or Squeeze Theorem),we conclude that:$\lim _{x \rightarrow \infty} \frac{f(6x)}{f(3x)} = 1$.

Let $f: R^{+} \rightarrow R$ be an increasing function such that $f(x) > 0$ for all $x$. If $\lim _{x \rightarrow \infty} \frac{f(9 x)}{f(3 x)}=1$,then $\lim _{x \rightarrow \infty} \frac{f(6 x)}{f(3 x)}=$

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