Evaluate $\lim _{x \rightarrow 0}\left(\frac{1}{\sqrt{x}}-\frac{\exp (-x)}{\sqrt{x}}\right)$

Question

According to WolframAlpha

$$\lim _{x \rightarrow 0}\left(\frac{1}{\sqrt{x}}-\frac{\exp (-x)}{\sqrt{x}}\right)=0$$

I do not understand this result since from me

$$\lim _{x \rightarrow 0}\left(\frac{1}{\sqrt{x}}-\frac{\exp (-x)}{\sqrt{x}}\right)= \lim _{x \rightarrow 0}\frac{1}{\sqrt{x}} - \lim _{x \rightarrow 0}\frac{\exp (-x)}{\sqrt{x}} = \infty - \infty$$

Answer 1

$\infty - \infty$ is an indeterminate form it does not mean anything, you need to use the definition of a limit or find a little trick :): $$\frac{1}{\sqrt{x}} - \frac{e^{-x}}{\sqrt{x}}= \frac{1-e^{-x}}{\sqrt{x}} = \sqrt{x} \cdot \frac{e^{-0} - e^{-x}}{x}$$ Since $\frac{e^{-x} - e^{-0}}{x} \to \frac{d}{dx}(e^{-x})(0) = -e^{-0} = -1$, by product of the limits, you get: $$\sqrt{x} \cdot \frac{e^{-0} - e^{-x}}{x} \to 0$$

Answer 2

$$\lim _{x \rightarrow 0}\left(\frac{1}{x^{0.5}}-\frac{\exp (-x)}{\sqrt{x}}\right)$$ $$=\lim_{x\to0}\left(\frac{1-e^{-x}}{\sqrt{x}}\right)$$ Recall that $$e^x=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\cdots$$ $$\implies 1-e^{-x}=x-\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$$ So our question becomes $$\lim_{x\to0}\frac{x-\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots}{\sqrt x}$$ Now apply L Hopital's Rule $$=\lim_{x\to0}\frac{1-\frac{2x}{2!}+\frac{3x^2}{3!}+\cdots}{\frac{1}{2\sqrt x}}$$ $$=\lim_{x\to0}\left(1-\frac{2x}{2!}+\frac{3x^2}{3!}+\cdots\right)2\sqrt x$$ $$=0$$

Answer 3

You can perhaps refer to a pre-existing answer on this site to clarify on your confusion: Can the difference of 2 undefined limits be defined?

As to how your original limit equates to zero, I would suggest you to represent e^(-x) in it's power series form and then evaluate the limit.

Answer 4

We have that by standard limit $\frac{e^{x}-1}{x}\to 1$

$$\frac{1}{\sqrt x}-\frac{e^{-x}}{\sqrt{x}}=\frac{e^{-x}-1}{-x}\frac{x}{\sqrt{x}}=\frac{e^{-x}-1}{-x}\sqrt{x}\to 1\cdot 0 =0$$

or as an alternative by squeeze theorem since $e^{-x}\ge 1-x$

$$0\le \frac{1}{\sqrt x}-\frac{e^{-x}}{\sqrt{x}} \le \frac{1}{\sqrt x}-\frac{1-x}{\sqrt{x}}=\frac{x}{\sqrt{x}}=\sqrt x\to 0 $$

---

Note that when by substitution we find an indeterminate form, as the one in hand $\infty-\infty$ the result can any and we need some more work and manipulation to get rid of it, for example as $x\to 0^+$

• $\frac1{x^2}-\frac1{x} =\frac1x\left(\frac1x-1\right) \to\infty\cdot \infty$

• $\frac{1+ax}{x}-\frac1{x} =a$

• $\frac1{x}-\frac1{x^2} =\frac1x\left(1-\frac1x\right) \to\infty\cdot -\infty=-\infty$

---

Refer also to

• What really is an indeterminate form?
• What do Indeterminate Forms mean?
• Undetermined or indeterminate forms: $\frac{0}{0}, \frac{\infty}{\infty}, 0\cdot\infty, 1^\infty, 0^0, +\infty-\infty$
• Why $ 0^{\infty} $ isn't indeterminate form?