I'm helping a friend with preparing an answer key for her calculus class, and I'm getting stuck with how to do the following limit.


According to Mathematica and Wolfram, the limit does not exist. However, the left hand limit does approach 1, and the right hand limit approaches $e^4$. How can I show these results analytically?

Thanks in advance.

  • $\begingroup$ If this is actually a question for a calculus class, then it would probably be a good idea for your friend to change this to a one-sided limit. $\endgroup$
    – user84413
    Mar 27, 2015 at 16:17

3 Answers 3


Let $t=\frac{1}{x}$; then $\displaystyle\lim_{x\to 0^{+}}\left(e^{8/x}+8x\right)^{x/2}=\lim_{t\to\infty}\left(e^{8t}+\frac{8}{t}\right)^{\frac{1}{2t}}$.

Since $\displaystyle\lim_{t\to\infty}\ln\left(e^{8t}+\frac{8}{t}\right)^{\frac{1}{2t}}=\lim_{t\to\infty}\frac{\ln\left(e^{8t}+\frac{8}{t}\right)}{2t}=\lim_{t\to\infty}\frac{8e^{8t}-\frac{8}{t^2}}{e^{8t}+\frac{8}{t}}\cdot\frac{1}{2}=$

$\hspace{.4 in}\displaystyle\lim_{t\to\infty}\frac{4e^{8t}-\frac{4}{t^2}}{e^{8t}+\frac{8}{t}}=\lim_{t\to\infty}\frac{4-\frac{4}{t^{2}e^{8t}}}{1+\frac{8}{te^{8t}}}=4,$

$\displaystyle \lim_{x\to 0^{+}}\left(e^{8/x}+8x\right)^{x/2}=\lim_{t\to\infty}\left(e^{8t}+\frac{8}{t}\right)^{\frac{1}{2t}}=e^4. $

Notice that $\displaystyle\lim_{x\to 0^{-}}\left(e^{8/x}+8x\right)^{x/2}$ is undefined, since $e^{8/x}+8x<0$ for $x<0$.

  • $\begingroup$ I think you divided by 2 one step too late with your L'Hopital. Otherwise it looks good. Additionally, I think that the limit is defined as it approaches from the other direction, it is just complex. $\endgroup$ Mar 27, 2015 at 0:39
  • $\begingroup$ I'm not sure I understand your comment about dividing by 2 one step too late; could you please explain what you mean by that. $\endgroup$
    – user84413
    Mar 27, 2015 at 15:07
  • $\begingroup$ The third expression after you write "since" where you perform L'Hopital, I feel that you're missing a 2 in the denominator. $\endgroup$ Mar 27, 2015 at 15:51
  • $\begingroup$ Thanks for your reply - - instead of dividing by 2, I multiplied by 1/2 (to make it easier to read). $\endgroup$
    – user84413
    Mar 27, 2015 at 16:10
  • $\begingroup$ Oh! Never mind. I was reading on my phone and it cut off the multiplication by 1/2. I'm silly $\endgroup$ Mar 27, 2015 at 16:54

Evaluate the limit from both sides:

Positive side

$$\lim_{x \to 0^+}\left(e^\frac{8}{x}+8x\right)^\frac{x}{2}$$

Piggybacking off of Tomi's suggestion, let $p = \left(e^\frac{8}{x}+8x\right)^\frac{x}{2}$ such that

$$\ln(p) = \frac{x}{2} \ln\left(e^\frac{8}{x}+8x\right)$$

As $x \to 0$, $8x \to 0$. However, $e^\frac{8}{x}$ blows up towards positive infinity. In our limit we can cancel out the 8x as it is a very small term and we are left with

$$\lim_{x\to 0^+}\left(\ln(p)\right) = \frac{x}{2} \ln\left(e^\frac{8}{x}\right) = \frac{x}{2}\frac{8}{x} = 4$$

$$\lim_{x\to 0^+}(p) = e^4$$

Negative side

We return again to $p$. For $x\to0$, we still have $8x\to0$, but now $e^{8/x}\to 0 $ much more rapidly than $8x$ because the exponent $\frac{8}{x}\to -\infty$ for $x\to 0^-$.

Thus we are left with

$$(8x)^\frac{x}{2} = 8^\frac{x}{2}x^\frac{x}{2}$$

This expression is continuous everywhere, but is complex for $x<0$ and real for $x\ge0$. Thankfully, because it is continuous everywhere, its positive limit at 0 is equal to its negative limit at 0, hence we can implement


$$=\lim_{x\to0^+}\exp \left(\ln(8^\frac{x}{2}x^\frac{x}{2})\right)$$

$$=\exp \left(\lim_{x\to0^+}\ln(8^\frac{x}{2})+\lim_{x\to0^+}\ln(x^\frac{x}{2})\right)$$

$$=\exp \left(\lim_{x\to0^+}\frac{x}{2}\ln(8)+\lim_{x\to0^+}\frac{x}{2}\ln(x)\right)$$

It is easy to demonstrate that $$\lim_{x\to0^+}\frac{x}{2}\ln(8)=0$$

However for $$\lim_{x\to0^+}\frac{x}{2}\ln(x)$$ we apply L'Hopital's rule.

$$\lim_{x\to0^+}\frac{x}{2}\ln(x)$$ $$=\frac{1}{2}\lim_{x\to0^+}\frac{\ln(x)}{1/x}$$ $$=\frac{1}{2}\lim_{x\to0^+}\frac{1/x}{-1/x^2}$$ $$=\frac{1}{2}\lim_{x\to0^+}(-x)$$ $$=0$$

Plugging each result back in,

$$\lim_{x\to0^-}8^\frac{x}{2}x^\frac{x}{2} = \exp(0) = 1$$

Putting it together.


$$\lim_{x\to0^+}\left(e^\frac{8}{x}+8x\right)^\frac{x}{2} = 1 \neq \lim_{x\to 0^+}\left(e^\frac{8}{x}+8x\right)^\frac{x}{2} = e^4$$ the limit


does not exist.


Try this:

Let $p=\left(e^\frac{8}{x}+8x\right)^\frac{x}{2}$

Then $ln(p)=\frac{x}{2} ln\left(e^\frac{8}{x}+8x\right)$

  • $\begingroup$ It seems to work for the positive side of you say that the 8x approaches zero but the $e^\frac{8}{x}$ blows up. Then because ln and exp are inverse functions for positive numbers, we are set. But I am still stuck on negatives. $\endgroup$ Mar 26, 2015 at 23:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .