Why limit for this function exists as $x$ approaches $0$?

Question

$$f(x) = \frac{\sqrt{x^{2}+100} - 10}{x^{2}} $$

As $$\lim_{x \to 0.01} f(x)=0.05$$ but, as we keep decreasing the value of $x$ to $0$, $$\lim_{x \to 0.000001} f(x) = 0.000000$$ So why this function's Limit exist?

Answer 1

Using enough accuracy in intermediate calculations,

$$\frac{\sqrt{0.000001^2+100}-10}{0.000001^2}\\=0.0499999999999998750000000000006249999999999960937500\cdots$$

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The expression is numerically unstable because of catastrophic cancellation. But you restore stability by using

$$\frac{\sqrt{x^{2}+100} - 10}{x^{2}}=\frac{{\left(\sqrt{x^{2}+100} -10\right)\left(\sqrt{x^{2}+100} +10\right)}} {x^2\left(\sqrt{x^{2}+100} +10\right)} =\frac1{\sqrt{x^{2}+100} +10} .$$

At the same time, it becomes obvious that the limit is $$\frac1{10+10}.$$

Answer 2

Note that

$$\lim_{x\to 0}\frac{\sqrt{x^{2}+100} - 10}{x^{2}}=\lim_{t\to 0^+}\frac{f(t)-f(0)}{t-0}=f'(0)$$

with

$$f(t)=\sqrt{t+100}\implies f'(t)=\frac{1}{2\sqrt{t+100}}\implies f'(0)=\frac1{20}=0.05$$

Maybe in your second evaluation there is a numerical problem in calculation.

Answer 3

Without using the definition of the derivate: $$\frac{\sqrt{x^2+100}-10}{x^2}=$$ Let's multiply the top and the bottom by $\sqrt{x^2+100}+10$ to get rid of the square root in the top: $$\frac{\sqrt{x^2+100}-10}{x^2}\frac{\sqrt{x^2+100}+10}{\sqrt{x^2+100}+10}=$$ $$\frac{\left(\sqrt{x^2+100}-10\right)\left(\sqrt{x^2+100}+10\right)}{x^2\left(\sqrt{x^2+100}+10\right)}=$$ $$\frac{\left(\sqrt{x^2+100}\right)^2-(10)^2}{x^2\left(\sqrt{x^2+100}+10\right)}=$$ $$\frac{x^2+100-100}{x^2\left(\sqrt{x^2+100}+10\right)}=$$ $$\frac{x^2}{x^2\left(\sqrt{x^2+100}+10\right)}=$$ $$\frac{1}{\sqrt{x^2+100}+10}\to \frac{1}{\sqrt{100}+10}=\frac{1}{20}$$ Or you can also use Taylor-series: $$\frac{\sqrt{x^2+100}-10}{x^2}=$$ $$\frac{10}{x^2}\left(\sqrt{1+\frac{x^2}{100}}-1\right)=$$ $$\frac{10}{x^2}\left(1+\frac{x^2}{200}+O(x^4)-1\right)=$$ $$\frac{10}{x^2}\left(\frac{x^2}{200}+O(x^4)\right)=$$ $$\frac{1}{20}+O(x^2)\to\frac{1}{20}$$