Limit with ln without L'Hôpital's rule

Question

This is an extra question in my maths assignment. Quite easy with L'Hôpital's, but I can't do it without the rule.

Answer 1

Observe that $$\frac{4x+5}{4x-5}=\frac{4x-5+10}{4x-5}=1+\frac{10}{4x-5}.$$ Then \begin{align*} \left(\frac{4x+5}{4x-5}\right)^x&=\exp\left\{x\log\!\left(1+\frac{10}{4x-5}\right)\right\}\\[.4em]&=\exp\left\{x\left(\frac{10}{4x-5}+o\!\left(\frac1x\right)\right)\right\}\\[.4em] &=\exp\left(\frac{10x}{4x-5}\right)\Bigl(1+o(1)\Bigr), \end{align*} which shows that $$\left(\frac{4x+5}{4x-5}\right)^x\xrightarrow[x\to\infty]{}\:\mathrm e^{\frac{10}4}=\sqrt{\mathrm e^5}.$$

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Other (more stupid) way. \begin{align*}\left(\frac{4x+5}{4x-5}\right)^x&=\left[\left(1+\frac{10}{4x-5}\right)^{\!\frac{4x-5}{10}}\right]^{\frac{10}4}\cdot\left(1+\frac{10}{4x-5}\right)^{\!\frac54}\\[.4em]&\xrightarrow[x\to\infty]{}\:\bigl[\mathrm e\bigr]^{\frac{10}4}\cdot1=\sqrt{\mathrm e^5}.\end{align*}