I have attempted to evaluate the limit $\lim _{x\to \infty }\left(\frac{1}{\sqrt{x-1}}\right)$. I understand, intuitively, that the limit is equal to zero, but when trying to show it, something strange happened. I've tried to evaluate the limit based on a technique that uses the fact that $\lim _{x\to \infty }\left(\frac{1}{x^n}\right)\:=\:0,\:n>0$, and divided both the numerator and the denominator of the fraction by $x$:
\begin{align}
& \lim _{x\to \infty }\left(\frac{1}{\sqrt{x-1}}\right)=\lim _{x\to \infty } \left(\frac{\frac{1}{x}}{\frac{\sqrt{x-1}}{x}}\right) =\lim_{x\to \:\infty} \left(\frac{\frac{1}{x}}{\frac{\sqrt{x-1}}{\sqrt{x^2}}}\right) =\lim_{x\to \:\:\infty \:\:}\left(\frac{\frac{1}{x}}{\sqrt{\frac{x-1}{x^2}}}\right) \\[10pt]
= {} & \lim_{x\to \:\:\:\infty \:\:\:} \left(\frac{\frac{1}{x}}{\sqrt{\frac{1}{x}-\frac{1}{x^2}}}\right)=\frac{\frac 1 \infty}{\sqrt{\frac 1 \infty -\frac{1}{\infty ^2}}}=\frac{0}{\sqrt{0-0}}=\frac{0}{0}= \text{indeterminate}
\end{align}
Moreover, when I divided the numerator and the denominator by $\sqrt{x}$, I've actually gotten the correct result:
$$\lim _{x\to \infty }\left(\frac{1}{\sqrt{x-1}}\right)=\lim _{x\to \infty } \left(\frac{\frac{1}{\sqrt{x}}}{\frac{\sqrt{x-1}}{\sqrt{x}}}\right)=\lim _{x\to \infty }\left(\frac{\frac{1}{\sqrt{x}}}{\sqrt{\frac{x-1}{x}}}\right)=\lim_{x\to \:\infty \:}\left(\frac{\frac{1}{\sqrt{x}}}{\sqrt{1-\frac{1}{x}}} \right) = \frac{\frac{1}{\sqrt{\infty }}}{\sqrt{1-\frac{1}{\infty }}}=\frac{0}{\sqrt{1-0}} = \frac{0}{1}=0$$
Why is there a difference in the results? Why does dividing by $\sqrt{x}$ is allowed, but not by dividing by $x$, or have I done an arithmetic/algebraic mistake that caused the result to be wrong?
Help will be much appreciated 🙂
Best Answer
The problem is right here:
$$\frac{0}{0}=\text{undefined}$$
The error is precisely the same as the one shown below, which is somewhat simpler: $$\lim_{u\to\infty} \frac{\sqrt{u}}{u}=\lim_{u\to\infty}\frac{\frac{1}{u}}{\frac{1}{\sqrt{u}}}=\frac{0}{0}$$
When you have a fraction of the form $\frac{a}{b}$ where both $a\to0$ and $b\to 0$, the limit of the ratio depends entirely on how fast the numerator and the denominator go to $0$. If the numerator shrinks more rapidly than the denominator, the ratio goes to $0$; if the denominator shrinks more rapidly than the numerator, the ratio goes to $\infty$; if they shrink at roughly the same pace, the ratio can converge to any value. This is in fact the whole principle behind derivatives: when we compute a quantity like $$\lim_{u\to 0}\frac{f(x+u)-f(x)}{u}$$ we generally don't get an undefined result, because the expression isn't actually $\frac{0}{0}$; rather it's a ratio of two quantities that separately go to $0$, but whose ratio converges to something specific.