# Evaluating the limit $\lim_{x \to 0}\left(x+e^{\frac{x}{3}}\right){}^{\!\frac{3}{x}}$

$$y=\left(x+e^{\frac{x}{3}}\right)^{\frac{3}{x}}$$

I'm looking at this equation, and need to solve for the limit as $\to 0$. I've graphed it, and know the solution is $e^4$, but am clueless as to the steps to actually solve this.

(Note, I am an adult working as a math aide in a high school. I help students at Algebra, Trig, Geometry, and Intro to Calculus. 35 years out of HS myself, there are clearly some things I need to brush up on. i.e. I know my limits. Pun intended.)

Yes, L'Hopital is fine. The student is in a BC calc class)

• Are you allowed to use L'Hopital's rule? – Miguel Feb 12 '16 at 13:44
• Yes, edited my answer for background. – JoeTaxpayer Feb 12 '16 at 13:52
• Basically, a way to see wat is going on is to see the affine approximation of $e^x$ around $0$: $$e^u \simeq e^0 + (e^\prime)(0) x = 1 + x$$ (this can be made formal by Taylor approximations to order $1$, for instance). This implies that your quantity is roughly $\left(x+ 1+ \frac{x}{3}\right)^{3/x} = \left(1+ \frac{4x}{3}\right)^{3/x}$, where you recognize, setting $t = \frac{3}{x}\to \infty$, the limit $$\left(1+\frac{4}{t}\right)^t \xrightarrow[t\to\infty]{} e^4.$$ The only key is to make this first approximation $\simeq$ rigorous. (NB: comment copied to my answer) – Clement C. Feb 12 '16 at 13:56
• 'Pun intended' +1 – Em. Feb 12 '16 at 13:57
• It may or may not make a difference, but you need to consider whether you mean $x\to 0$, $x\to0^+$, or $x\to0^-$ since there is a $\frac1x$ involved. – MPW Feb 12 '16 at 14:39

$$y=\left(x+e^{\frac{x}{3}}\right)^{\frac{3}{x}}$$ $$\log(y)={\frac{3}{x}}\log\left(x+e^{\frac{x}{3}}\right)$$ Now, using that, close to $t=0$, $e^t\simeq 1+t$.

So $$\log(y)\simeq {\frac{3}{x}}\log\left(x+1+{\frac{x}{3}}\right)={\frac{3}{x}}\log\left(1+{\frac{4x}{3}}\right)$$ Now, using that, close to $t=0$ $\log(1+t)\sim t$, $$\log(y)\simeq {\frac{3}{x}}{\frac{4x}{3}}=4$$

• log(1+t)∼t was key. That's perfect, much thanks. I started playing with log, and missed that.... – JoeTaxpayer Feb 12 '16 at 13:55
• @JoeTaxpayer. You are welcome ! When you have a power (function of $x$), it is a very common trick. Don't forget it; it is very useful and you will need it a lot. Cheers. – Claude Leibovici Feb 12 '16 at 13:57
• @ClaudeLeibovici This will look like splitting hair, but the use of $\sim$ (equivalent) for more than one term in the RHS is misleading here (and technically incorrect). By definition, you could say $e^t \sim_{t\to 0} 1 +t^{0.0001}$, this would also be correct, as $$\frac{e^t}{1 +t^{0.0001}} \to 1$$ as well. For more than one term, using $o(\cdot)$ or $O(\cdot)$ is much less prone to these issues. – Clement C. Feb 12 '16 at 14:00
• @ClementC.. Thanks for pointing it out ! I shall fix it now. Being almost blind makes that I have often mistakes of this kind. Cheers. – Claude Leibovici Feb 12 '16 at 14:02
• @ClaudeLeibovici No worries! (My math professors made sure we memorized that subtlety, sometimes a bit harshly.) – Clement C. Feb 12 '16 at 14:04

Intuition: A way to see what is going on is to see the affine approximation of $e^x$ around $0$: $$e^u \simeq e^0 + (e^\prime)(0) x = 1 + x$$ (this can be made formal by Taylor approximations to order $1$, for instance). This implies that your quantity is roughly $\left(x+ 1+ \frac{x}{3}\right)^{3/x} = \left(1+ \frac{4x}{3}\right)^{3/x}$, where you recognize, setting $t = \frac{3}{x}\to \infty$, the limit $$\left(1+\frac{4}{t}\right)^t \xrightarrow[t\to\infty]{} e^4.$$ The only key is to make this first approximation $\simeq$ rigorous, which is done below.

An approach based on Taylor expansions: (but which requires no knowledge of them besides the Landau notation $o(\cdot)$ — justifying what is needed as we go)

Start (as often when you have both a base and an exponent depending on $x$) by rewriting it in exponential form: $$\left(x+e^{\frac{x}{3}}\right)^\frac{3}{x} = e^{\frac{3}{x}\ln\left(x+e^{\frac{x}{3}}\right)}$$

Now, when $u\to 0$, we have $\frac{e^u-1}{u}\to \exp^\prime 0 = e^0 = 1$, so that $e^u = 1+u + o(u)$; which gives $$x+e^{\frac{x}{3}} = x+1+ \frac{x}{3} + o(x) = 1+\frac{4}{3}x.$$

Similarly, since $\frac{\ln(1+u)}{u}\xrightarrow[u\to 0]{} 1$, we have $\ln(1+u) = u + o(u)$. Combining the two, we get $$\ln\left(x+e^{\frac{x}{3}}\right) = \ln\left(1+\frac{4}{3}x\right) = \frac{4}{3}x + o(x).$$

Putting it together, $$\frac{3}{x}\ln\left(x+e^{\frac{x}{3}}\right) = \frac{3}{x}\left(\frac{4}{3}x + o(x)\right) = 4 + o(1) \xrightarrow[x\to 0]{} 4$$ and, by continuity of $\exp$, $$e^{\frac{3}{x}\ln\left(x+e^{\frac{x}{3}}\right)} \xrightarrow[x\to 0]{} e^4.$$

Hint Provided that $\lim_{x \to 0} \log y$ exists, by continuity we have $$e^{\lim_{x \to 0} \log y} = \lim_{x \to 0} e^{\log y} = \lim_{x \to 0} y,$$ and on the other hand, $$\log y = 3 \cdot \frac{\log(x + e^{x / 3})}{x}.$$ Now, we can evaluate the limit of $\log y$ as $x \to 0$ by (1) applying L'Hopital's Rule, or (2) using some elementary Taylor series expansions to conclude that that $$x + e^{x / 3} = 1 + \frac{4}{3} x + O(x^2)$$ and hence $$\log(x + e^{x / 3}) = \frac{4}{3} x + O(x^2) .$$

If $L$ is the desired limit then \begin{align} \log L &= \log\left\{\lim_{x \to 0}\left(x + e^{x/3}\right)^{3/x}\right\}\notag\\ &= \lim_{x \to 0}\log\left(x + e^{x/3}\right)^{3/x}\text{ (via continuity of log)}\notag\\ &= \lim_{x \to 0}\frac{3}{x}\cdot\log\left(x + e^{x/3}\right)\notag\\ &= \lim_{x \to 0}\frac{3}{x}\cdot\log\left(1 + x + e^{x/3} - 1\right)\notag\\ &= \lim_{x \to 0}\frac{3}{x}\cdot (x + e^{x/3} - 1)\cdot\frac{\log\left(1 + x + e^{x/3} - 1\right)}{x + e^{x/3} - 1}\notag\\ &= \lim_{x \to 0}\frac{3}{x}\cdot (x + e^{x/3} - 1)\cdot\lim_{t \to 0}\frac{\log(1 + t)}{t}\text{ (putting }t = x + e^{x/3} - 1)\notag\\ &= \lim_{x \to 0}\left(3 + \frac{e^{x/3} - 1}{x/3}\right)\notag\\ &= 3 + \lim_{t \to 0}\frac{e^{t} - 1}{t}\text{ (putting }t = x/3)\notag\\ &= 3 + 1 = 4\notag \end{align} and hence $L = e^{4}$.