Easier method to prove limit by epsilon delta definition.

Question

supposed I wish to prove this limit via the epsilon delta definition of limits:

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

This means I have to show that:

$$\bigg|\frac{1}{\sqrt{x^2+3}} - \frac{1}{2}\bigg| < \epsilon$$ whenever $0<\big|x +1\big|<\delta$.

How can I proceed to choose $\delta$ in a way such that my proof will be simple? So far I have tried to work "backwards" but ended up all messy because of the squareroot.

Any hint or help would be appreciated. Thanks!

Answer 1

Let $\varepsilon > 0$; then $$ \begin{split} \left| \frac{1}{(x^{2}+3)^{1/2}} - \frac{1}{2}\right| &= \left| \frac{2 - (x^{2}+3)^{1/2}}{2(x^{2}+3)^{1/2}}\right| \\ &= \frac{|x-1||x+1|}{4(x^{2}+3)^{1/2} + 2(x^{2}+3)} \\ &< |x-1||x+1| \\ &< |x+1| \\ &< \varepsilon \end{split}$$ if $|x+1| < 3$ and $< \varepsilon$, so taking $\delta := \min \{ 3, \varepsilon \}$ suffices.

Answer 2

$$\left|\frac{1}{\sqrt{x^2+3}}-\frac{1}{2}\right|=\frac{|2-\sqrt{x^2+3}|}{2\sqrt{x^2+3}}=\frac{\left|2-|x|\sqrt{1+\frac{3}{x^2}}\right|}{2\sqrt{x^2+3}}$$

Take $-2\leq x\leq 0$, then

you can easily show that $$\left|2-|x|\sqrt{1+\frac{3}{x^2}}\right|\leq |x+1|$$

and conclude.