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?

  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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

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

  • $\begingroup$ 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
  • 4
    $\begingroup$ 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

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:

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.