Trouble finding approach for limit

Question

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?

Thanks in advance.

Answer 1

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. $

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

Answer 2

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.

Answer 3

Try this:

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

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