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I know that $$ \lim_{x \to 0+} \frac{ \sqrt{ x+1 } }{x} = +\infty. $$

Now how to find $$ \lim_{x \to 0- } \frac{ \sqrt{ x+1 } }{x} = +\infty? $$

This is Prob. 5 (e), Sec. 4.3, in the book Introduction To Real Analysis by Robert G. Bartle and Donald R. Sherbert, 4th edition.

My Attempt:

The function $f$ given by $$ f(x) \colon= \frac{ \sqrt{ x+1 } }{x} $$ is defined only for real numbers $x \geq -1$. Moreover, $f(x) < 0$ for $x < 0$.

If $-1 < x < 0$, then $$ 0 < \sqrt{ x+1} < 1, $$ and so $$ \frac{1}{x} < \frac{ \sqrt{ x+1} }{x} < 0. \tag{1} $$

Also if $-1 < x < 0$, then $$ -\frac{1}{x} > 1, $$ and as $\sqrt{x+1} > 0$, so $$ - \frac{ \sqrt{ x+1} }{x} > \sqrt{x+1}, $$ and hence $$ \frac{ \sqrt{ x+1} }{x} < - \sqrt{x+1}. $$

What to do here?

I've really no clue of what to do. I would like to be able to majorise this function by one tending to $-\infty$ as $x \to 0-$, or I would like to bound this function between two functions that both have the same limit as $x \to 0-$.

Or, does the limit exist at all in the extended real number system $\mathbb{R} \cup \left\{ \ \pm \infty \ \right\}$?

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  • $\begingroup$ I think you are wrong, it should be $-\infty$. $\endgroup$ Aug 6, 2018 at 15:40
  • $\begingroup$ Intuitively numerator tends to 1 and numerator to 0 from the negative side therefore the LHS limit is $-\infty$. To show that rigoursly we can use for example squeeze theorem. $\endgroup$
    – user
    Aug 6, 2018 at 15:45

3 Answers 3

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We have that as $x\to 0^+$

$$\frac{ \sqrt{ x+1 } }{x}\ge \frac 1 x \to \infty$$

and as $x\to 0^-$ let $y=-x \to 0^+$ eventually as $\sqrt{ 1-y }\ge \frac12$

$$\frac{ \sqrt{ x+1 } }{x}=-\frac{ \sqrt{ 1-y } }{y}\le -\frac 1 {2y} \to -\infty$$

therefore the limit doesn't exist but the one side limits exist in $\mathbb{R} \cup \left\{ \ \pm \infty \ \right\}$.

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The stated result isn't right. $1/x \to -\infty$ as $x\to 0^-$. The numerator is bounded and positive near $x=0$, so the answer remains the same. A full proof: Since $\sqrt{1+x}$ is increasing in $x$,

$$ \frac{\sqrt{1+x}}{x} ≤ \frac{\sqrt{1+0}}{x} = \frac1x \to -\infty \quad (x\to 0^-) $$

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$$ \frac{\sqrt{x + 1}}{x} =\frac{x + 1}{x\sqrt{x + 1}} =\frac{1}{\sqrt{x + 1}} + \frac{1}{x\sqrt{x + 1}} \to -\infty \quad \text{as} \quad x \to 0^- $$

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  • $\begingroup$ @stevengregory thank you very much for such a beautiful answer! I should be cross with myself for not being able to figure it out on my own! $\endgroup$ Aug 6, 2018 at 17:33

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