# Finding a limit results in division by 0.

$$\lim_{x\to0}\dfrac{\sqrt{1+x}-1}{x^2}.$$ Tried multiplying by $\sqrt{1+x}+1$ but got $1/(x(\sqrt{1+x}+1))$ where substitution results in $1/0$ which is illegal (or maybe $1/0$ is Infinity?). Second approach is to divide by $x^2$ which leads to the $\sqrt{1/x^2 +1)}- 1/x^2$. How to solve it?

• excuse but if you set your exact example Nov 15, 2013 at 15:19

Your first approach was spot on, though your conclusion is off a bit:

Multiplying the numerator and denominator by the conjugate, as you suggest, and canceling a common factor of $x$ from the numerator and denominator gives us, as you found, $$\lim_{x\to 0} \dfrac{1}{x(\sqrt{1+ x} + 1)}$$ which at first glance seems to result in a "form" of $\frac 10$. This is not "illegal" when we're evaluating limits. However, in this case that the limit does not exist, since as $x\to 0^-$, $f(x) \to -\infty$, whereas as $x\to 0^+$, $f(x)\to +\infty$.

Conclusion: In this case, since the left-side and right-side limits to not agree, the limit does not exist.

Note: It's not always the case that a limit of the form $\frac 10$ does not exist, in the sense that the limits from the left, and from the right, do not agree. For example, $\lim_{x \to 0} \dfrac 1{x^2} \to +\infty$, since as $x$ approaches $0$ from the right and from the left, the denominator is always positive (unlike your posted limit), and the denominator grows increasingly and incredibly small compared to $1$ in the numerator, and as a result, the limit approaches (positive) infinity (the value of the function explodes) as $x \to 0$.

The non-existence of a limit, or a limit approaching infinity as $x$ approaches a zero from both directions, is not the same as (or because of) "division by zero". Evaluating a limit as $x \to 0$ is not the same as evaluating a function $f(x)$ at zero. Here, we are interested in what's happening as $x$ approaches zero from the right and from the left, and not what's happening exactly at zero.

• Thanks! Wolfram confirms You conclusion. But in my exercise book the answer is "Infinity". Is that not accurate answer? Nov 15, 2013 at 16:42
• No, it's not accurate - since x, as $x \to 0$ from the left is negative, and so the limit as $x \to 0$ from the left is $-\infty$. From the right, x remains positive as $x\to 0$, so the limit as $x \to 0$ from the right is $+\infty$. When limits from either side of 0 do not agree, the limit does not exist. I would write out an answer that notes the different results (approaching $0$ from left compared to the right), to explain why the limit does not exist. That way, you're covering your bases. Nov 15, 2013 at 16:45
• @Amzoti Thanks for the support! :-) Nov 16, 2013 at 1:57

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

If $x\rightarrow 0^- \Rightarrow \frac{1}{x(\sqrt{x+1}+1)}\rightarrow -\infty$, and

$x\rightarrow 0^+ \Rightarrow \frac{1}{x(\sqrt{x+1}+1)}\rightarrow +\infty$

• How is it helpful to just completely solve the problem for the OP?
– LASV
Nov 15, 2013 at 15:48

knowing $\lim_{x \to 0}\frac{1}{x}$ does not exist,while $\lim_{x \to 0}\frac{1}{x^2}=\infty$ we compute

$\lim_{x \to 0}\frac{\sqrt{x+1}-1}{x^2}=\lim_{x \to 0}\sqrt{1+\frac{1}{x}}-\frac{1}{x^2}$ and we see that the limit does not exists

For $x\to 0, \sqrt{1+x}\sim1+\frac{1}{2}x-\frac{1}{8}x^2$ so that for $x\to 0^{\pm}$ $$\frac{\sqrt{1+x}-1}{x^2}\sim \frac{1}{2x}-\frac{1}{8}\to\pm\infty$$