# How to find the limit using L'Hôspital's Rule

Find the limit. Use L'Hôspital's Rule if appropriate. If there is a more elementary method, consider using it.

Find the limit as x approaches infinity of $(\frac{5x-3}{5x+4})^{5x+1}$

My first thought would be to take the natural log and rewrite it as the exponent times the fraction rewritten as adding two natural log functions, but I don't know how that would help me. Can anyone help?

• Try to rewrite into something like this: $\displaystyle lim_{x\to \infty} \left(1 + \dfrac{a}{5x+4} \right)^{(5x+4)\cdot f(x)}$ where you can use the limit for $e$ – FormerMath Oct 10 '16 at 21:59
• The edit by @BirlantisEscheatvc is unnecessary because L'Hospital is a correct spelling of the name. See here for example for a brief explanation. – 6005 Oct 10 '16 at 22:10
• @6005, my bad, I am a little rusty. I changed everything back to normal. – 关一骏 Oct 10 '16 at 22:17
• You posted three questions with very similar titles in less than one hour. – egreg Oct 10 '16 at 22:18
• @Luis Try to prepend a backslash to 'lim', so it becomes a $\LaTeX$ symbol and gets rendered in upright font: \lim → $\lim_{x\to\infty}$ instead of italics looking like a multiplicaton: lim → $lim =(?) l\cdot i\cdot m$. – CiaPan Oct 11 '16 at 7:21

$$\lim_{x\rightarrow \infty }\left(\frac{5x-3}{5x+4}\right)^{5x+1}=\lim_{x\rightarrow \infty }\left(1-\frac{7}{5x+4}\right)^{5x+1}$$ $$=\lim_{x\rightarrow \infty }\left(1-\frac{7}{5x+4}\right)^{-3}\lim_{x\rightarrow \infty }\left(1-\frac{7}{5x+4}\right)^{5x+4}$$ $$=\left(1\right)\lim_{x\rightarrow \infty }\left(1-\frac{7}{5x+4}\right)^{5x+4}=\frac{1}{e^7}$$

• The answer you gave was incorrect. – Maggie Oct 10 '16 at 22:10
• Why...........? – E.H.E Oct 10 '16 at 22:23
• This looks just fine to me, although it didn't use l'Hospital. +1 – DonAntonio Oct 10 '16 at 22:26
• @DonAntonio thanks – E.H.E Oct 10 '16 at 22:29
• Wolfram Alpha is telling me this answer is right. – user361424 Oct 10 '16 at 23:18

Although not as pretty as just using the limit definition of $e$, you can calculate it via L'Hôpital's rule by first rewriting the limit as follows:

$$\lim_{x\rightarrow \infty}\left(\frac{5x-3}{5x+4}\right)^{5x+1} = \lim_{x\rightarrow \infty}\operatorname{exp}\left[(5x+1)\operatorname{ln}\left(\frac{5x-3}{5x+4}\right)\right] = \lim_{x\rightarrow \infty}\operatorname{exp}\left[\frac{\operatorname{ln}\left(\frac{5x-3}{5x+4}\right)}{\frac{1}{5x+1}}\right]\\ = \operatorname{exp}\left[\lim_{x\rightarrow \infty}\frac{\operatorname{ln}\left(\frac{5x-3}{5x+4}\right)}{\frac{1}{5x+1}}\right].$$

• Nice one still :) – mick Oct 10 '16 at 22:58