Finding a limit results in division by 0.

Question

$$ \lim_{x\to0}\dfrac{\sqrt{1+x}-1}{x^2}. $$ 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?

Answer 1

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.

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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.

Answer 2

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

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

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

Answer 3

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

Answer 4

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