# Left hand Limit at zero

$$\lim_{x\to 0^{-}} \frac{\sqrt{x+1}}{x} \quad (x>-1),$$

I am able to find the right hand limit at zero. As for $x>0, \frac{\sqrt{x+1}}{x}> \frac{1}{x}$ and since $\lim_{x\to 0^{+}}\frac{1}{x} = \infty$, we have $$\lim_{x\to 0^{+}} \frac{\sqrt{x+1}}{x}= \infty$$.

Thanks!

P.S: Please avoid using denominator goes to zero, numerator goes to infinity. I quote the definition for reference. If for any k $\in R$, $f(x)>k$ when x is in neighborhood of c, we say lim of f at c is infinity.

• In the first limit you have $0^-$; do you also want the left-sided limit or is that a typo? Also, what is your question exactly - to check your work? Commented Dec 8, 2017 at 14:37
• @StackTD So sorry! Thanks for noticing Commented Dec 8, 2017 at 14:39

For $-3/4<x<0$ you have $1/4<x+1<1$ and so $1/2<\sqrt{x+1}<1$, so $$\frac{\sqrt{x+1}}{x}<\frac{1}{2x}$$ Since $\lim_{x\to0^-}\frac{1}{2x}=-\infty$ you have your conclusion.

Since $x$ is negative in the left-hand limit, you have $x = -\sqrt{x^2}.$ Then

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

Inside the radical, as $x$ gets close to zero, the top goes to $1$ while the bottom is positive and goes to $0$. So the radicand goes to $+\infty$. So your limit is $-\sqrt{+\infty} = -\infty.$

By algebric rules for limits:

For $x>0$

$$\frac{\sqrt{x+1}}{x}=\frac1x\sqrt{x+1}=+\infty\cdot 1= +\infty$$

For $x<0$ set $y=-x$

$$-\frac{\sqrt{1-y}}{y}=-\frac1y\sqrt{1-y}=-\infty\cdot 1= -\infty$$

You can easily find the limits on both sides by observing that near $x=0$, the numerator is always positive (since it's a square root) so the sign only depends on the sign of the denominator, which is simply $x$; so we have: $$\lim_{x \to 0^-} \frac{\sqrt{x+1}}{x} = -\infty \quad \mbox{ and } \quad \lim_{x \to 0^+} \frac{\sqrt{x+1}}{x} = +\infty$$

• A rigorous proof? Commented Dec 8, 2017 at 14:49
• How rigorous do you want it? There's always the definition... :-) Please be more clear in what you ask or want to know. Commented Dec 8, 2017 at 14:50
• @yasir: this is as rigorous as your attempt in the question. If you want a typical $\epsilon$ proof then I think you should not have much issue doing that based on the arguments given in this answer. Commented Dec 8, 2017 at 15:00