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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?

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    $\begingroup$ 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 $e$ $\endgroup$ – FormerMath Oct 10 '16 at 21:59
  • $\begingroup$ The edit by @BirlantisEscheatvc is unnecessary because L'Hospital is a correct spelling of the name. See here for example for a brief explanation. $\endgroup$ – 6005 Oct 10 '16 at 22:10
  • $\begingroup$ @6005, my bad, I am a little rusty. I changed everything back to normal. $\endgroup$ – 关一骏 Oct 10 '16 at 22:17
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    $\begingroup$ You posted three questions with very similar titles in less than one hour. $\endgroup$ – egreg Oct 10 '16 at 22:18
  • $\begingroup$ @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 → $lim =(?) l\cdot i\cdot m$. $\endgroup$ – CiaPan Oct 11 '16 at 7:21
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$$\lim_{x\rightarrow \infty }\left(\frac{5x-3}{5x+4}\right)^{5x+1}=\lim_{x\rightarrow \infty }\left(1-\frac{7}{5x+4}\right)^{5x+1}$$ $$=\lim_{x\rightarrow \infty }\left(1-\frac{7}{5x+4}\right)^{-3}\lim_{x\rightarrow \infty }\left(1-\frac{7}{5x+4}\right)^{5x+4}$$ $$=\left(1\right)\lim_{x\rightarrow \infty }\left(1-\frac{7}{5x+4}\right)^{5x+4}=\frac{1}{e^7}$$

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

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

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  • $\begingroup$ Nice one still :) $\endgroup$ – mick Oct 10 '16 at 22:58

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