Calculate $\lim_{x \to 0^+} \frac{3x + \sqrt{x}}{\sqrt{1- e^{-2x}}}$

Question

I need to calculate: $$\displaystyle \lim_{x \to 0^+} \frac{3x + \sqrt{x}}{\sqrt{1- e^{-2x}}}$$

I looks like I need to use common limit: $$\displaystyle \lim_{x \to 0} \frac{e^x-1}{x} = 1$$

So I take following steps:

$$\displaystyle \lim_{x \to 0^+} \frac{3x + \sqrt{x}}{\sqrt{1- e^{-2x}}} = \displaystyle \lim_{x \to 0^+} \frac{-3x - \sqrt{x}}{\sqrt{e^{-2x} - 1}}$$

And I need to delete root in the denominator and make nominator equal to $-2x$. But I don't know how.

Answer 1

Note that\begin{align}\lim_{x\to0^+}\frac{\left(3x+\sqrt x\right)^2}{1-e^{-2x}}&=\lim_{x\to0^+}\frac{9x^2+6x\sqrt x+x}{1-e^{-2x}}\\&=-\lim_{x\to0^+}\frac x{e^{-2x}-1}\left(9x+6\sqrt x+1\right)\\&=-\frac1{\lim_{x\to0^+}\frac{e^{-2x}-1}x}\times\lim_{x\to0^+}\left(9x+6\sqrt x+1\right)\\&=-\frac1{-2}\times1\\&=\frac12.\end{align}Therefore, the limit that you're after is $\sqrt{1/2}$.

Answer 2

$$\lim_{x \to 0^+} \frac{3x + \sqrt{x}}{\sqrt{1- e^{-2x}}} =\lim_{x \to 0^+}\frac{\sqrt{x}}{\sqrt{1- e^{-2x}}} \cdot(3\sqrt{x} + 1) = \lim_{x \to 0^+} \sqrt{\frac{x}{1- e^{-2x}}} \cdot\lim_{x \to 0^+} (3\sqrt{x} + 1)$$

$$\lim_{x \to 0^+} \sqrt{\frac{x}{1- e^{-2x}}} = \lim_{x \to 0^+} \sqrt{\frac{-2x}{(2)( e^{-2x} -1)}}= \frac{1}{\sqrt{2}}$$

Answer 3

$$\sqrt{1-e^{-2 x}}\sim \sqrt{2} \sqrt{x};\;\text{as }x\to 0$$

$$\lim_{x \to 0^+} \frac{3x + \sqrt{x}}{\sqrt{2} \sqrt{x}}=\lim_{x \to 0^+}\left(\frac{3}{\sqrt 2}\sqrt x +\frac{1}{\sqrt 2} \right)=\frac{1}{\sqrt 2} $$