I'm trying to find $\lim\limits_{x \to \infty} \left (2 + \frac{3}{x}\right)^{5x}$ which should equal to $\infty$. But in the process of using $e^{\lim\limits_{x \to \infty} 5x\ln (2 + \frac{3}{x})}$, I got $\frac{15}{2}$ as an answer to the exponent. This is what I did.

Thank you.

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    $\begingroup$ $\log(2)\ne \infty$. And $\infty/0$ is not indeterminate any way, $\endgroup$ – Mark Viola Jan 27 '17 at 23:01
  • $\begingroup$ @Dr.MV I get it now. Thank you! $\endgroup$ – DaLemon Jan 27 '17 at 23:20

You make a mistake near the very beginning:


You mistakenly think that $\ln(2+\frac3x)\to\ln(1)=0$, but it actually approaches $\ln(2)$. Thus, your L'Hospital's is invalid. One should instead get


  • $\begingroup$ Mind the lazy notation at the end. $\endgroup$ – Simply Beautiful Art Jan 27 '17 at 23:04
  • $\begingroup$ That makes much more sense. Thank you so much!!! $\endgroup$ – DaLemon Jan 27 '17 at 23:07
  • $\begingroup$ @DaLemon No problem :D It is commonly the problem in an incorrectly done limit that the indeterminate form step was unfulfilled upon applying L'Hospital's rule, so keep that in mind. $\endgroup$ – Simply Beautiful Art Jan 27 '17 at 23:08
  • $\begingroup$ @DaLemon Since you seem to be new to this, I will point out that if you think an answer is good/interesting/useful, you can upvote by clicking the upwards arrow on the left side of an answer, and if you think an answer has successfully answered your question, you can click the green checkmark button below the arrows. $\endgroup$ – Simply Beautiful Art Jan 27 '17 at 23:12
  • $\begingroup$ Alrighty! I shall do that. Thank you again! $\endgroup$ – DaLemon Jan 27 '17 at 23:15

$$\begin{align}\lim _{x\to \infty }\left(\:5x\ln\:\left(2\:+\:\frac{3}{x}\right)\right) &= \lim _{t\to 0}\left(\:\frac{5}{t}\:\cdot \:\ln\left(2\:+\:3t\right)\right) \\&= \lim _{t\to 0}\left(\:\frac{5}{t}\:\cdot \:\left(\ln \:\left(2\right)+\frac{3}{2}t+o\left(t\right)\right)\right) \\&= \lim _{t\to 0}\left(\:\frac{5\ln\left(2\right)}{t}\:+\frac{15}{2}\right) \\&= \infty\end{align}$$

Is an approximation obtained by the Taylor formula to the first order....look this https://en.wikipedia.org/wiki/Taylor%27s_theorem
Taylor's formula: $$\sum _{n=0}^{\infty }\:\frac{f^{\left(n\right)}\left(x_0\right)}{n!}\left(x-x_0\right)^n$$

So in our case$(x_0 = 0)$: $$\ln(2+3x)=\ln \left(2\right)+\frac{\frac{3}{2}}{1!}x+\frac{-\frac{9}{4}}{2!}x^2+\frac{\frac{27}{4}}{3!}x^3+\frac{-\frac{243}{8}}{4!}x^4+\ldots $$ but it is enough to stop at the first order.

  • $\begingroup$ how do you get the part where it says ln(2) + 3/2t + o(t)? $\endgroup$ – DaLemon Jan 27 '17 at 23:27
  • $\begingroup$ @DaLemon i edited my answer $\endgroup$ – Amarildo Jan 27 '17 at 23:41
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    $\begingroup$ With Lagrange remainder bounds, one can remove your $\approx$ when you took the Taylor expansion and instead apply squeeze theorem if you wish to be rigorous. $\endgroup$ – Simply Beautiful Art Jan 28 '17 at 2:55

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