solving the limit of $(e^{2x}+x)^{1/x}$

Question

I tried to solve for the following limit:

$$\lim_{x\rightarrow \infty} (e^{2x}+x)^{1/x}$$ and I reached to the indeterminate form: $${4e^{2x}}\over {4e^{2x}}$$ if I plug in, I will get another indeterminate form!

Answer 1

Tricky solution: $$\begin{align} (e^{2x}+x)^{1/x}&=e^{2x/x}\left(1+\frac{1}{e^{2x}/x}\right)^{1/x}\\ &=e^2\left(\left(1+\frac{1}{e^{2x}/x}\right)^{e^{2x}/x}\right)^{e^{-2x}} \end{align}$$

Since $e^{2x}/x\to\infty$ and $(1+1/y)^y\to e$ as $y\to\infty$, and $e^{-2x}\to 0$, you get the limit is $e^2$.

Answer 2

I have no idea how you got $\lim\limits_{x\to\infty}\frac{4e^{2x}}{4e^{2x}}$, but as has been pointed out, that limit is easily evaluated: the fraction is identically $1$, so the limit is also $1$.

Let $L=\lim\limits_{x\to\infty}\left(e^{2x}+x\right)^{1/x}$; then

$$\ln L=\ln\lim_{x\to\infty}\left(e^{2x}+x\right)^{1/x}=\lim_{x\to\infty}\ln\left(e^{2x}+x\right)^{1/x}=\lim_{x\to\infty}\frac{\ln\left(e^{2x}+x\right)}x\;.$$

Now apply l’Hospital’s rule.

Answer 3

Another way, considering $$A=(e^{2x}+x)^{1/x}$$ Taking logarithms $$\log(A)=\frac 1x \log(e^{2x}+x)=\frac 1x \left(\log(e^{2x})+\log(1+\frac x {e^{2x}})\right)=\frac 1x \left(2x+\log(1+\frac x {e^{2x}})\right)$$ $$\log(A)=2+\frac 1x \log(1+\frac x {e^{2x}})\approx 2+\frac 1x \frac x {e^{2x}}=2+\frac 1 {e^{2x}}$$

Answer 4

$x>0\implies e^x>x \implies e^2=(e^{2 x})^{1/x}<(e^{2x}+x)^{1/x}< (2 e^{2x})^{1/x} =(e^2) (2^{1/x})$..... And $\lim_{x\to \infty}2^{1/x}=1.$

Answer 5

We simply have

$$(e^{2x}+x)^{1/x}=e^{2x/x}\left(1+\frac{x}{e^{2x}}\right)^{1/x}\to e^2\cdot1^0=e^2$$