# Trouble finding approach for limit

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.

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

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?

• 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. Mar 27, 2015 at 16:17

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 $x<0$.

• 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. Mar 27, 2015 at 0:39
• 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. Mar 27, 2015 at 15:07
• The third expression after you write "since" where you perform L'Hopital, I feel that you're missing a 2 in the denominator. Mar 27, 2015 at 15:51
• Thanks for your reply - - instead of dividing by 2, I multiplied by 1/2 (to make it easier to read). Mar 27, 2015 at 16:10
• Oh! Never mind. I was reading on my phone and it cut off the multiplication by 1/2. I'm silly 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

$$\lim_{x\to0^+}8^\frac{x}{2}x^\frac{x}{2}$$

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

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

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

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.

Because

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

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

does not exist.

Try this:

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

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

• 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. Mar 26, 2015 at 23:41