3
$\begingroup$

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 :)

$\endgroup$
  • 1
    $\begingroup$ In the first case, the conclusion should be "indeterminate": something more needs to be done. $\endgroup$ – Namaste Jul 4 '18 at 16:38
  • 1
    $\begingroup$ Think about it this way: $\lim_{x \to 0} x = 0$, but $\lim_{x \to 0} \frac{x^2}{x} = \frac{0}{0}$? Making limits indeterminate is easy. Determining them can be less easy. :-) $\endgroup$ – Theo Bendit Jul 4 '18 at 16:50
  • 2
    $\begingroup$ There's no difference in the results because in the first case you didn't get any result. You said that $0/0$ is indeterminate, but that's not a result: It does not attempt to say what the limit is. $\endgroup$ – Michael Hardy Jul 4 '18 at 17:16
  • $\begingroup$ No! You can't plug indeterminates into an expression! $\lim \frac {a_n}{b_n} \ne \frac {\lim a_n}{\lim b_n}$ unless $\lim a_n$ and $\lim b_n$ are both existent real numbers. $\endgroup$ – fleablood Jul 4 '18 at 17:59
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

$$ \lim_{x\,\to\,\infty} \frac 1 {\sqrt{x-1}} $$ Your making a simple thing incredibly complicated. You don't need to deal with any indeterminate form here. \begin{align} x & \to \infty \\[10pt] x-1 & \to \infty \\[10pt] \sqrt{x-1} & \to \infty \\[10pt] \frac 1 {\sqrt{x-1}} & \to 0 \end{align} Only when you have an indeterminate form do you need to do more than that. You changed it into something where you had an indeterminate form, but there was no need for that.

You did not get two different results; rather, with your first method you got no result at all. When you correctly identified $0/0$ as an indeterminate form, that's not a result; it does not say nor attempt to say what the limit is.

In many problems – in fact in all problems involving the definition of "derivative" – you will start with the indeterminate form $0/0,$ and the limit will often be some particular number.

$\endgroup$
1
$\begingroup$

You lose a lot of the nuances of a fraction when you deliberately multiply the numerator and denominator by 0. This is a mathematical anti-strategy that would sabotage your ability to solve most problems. I figured out in pre-algebra that you can't simply multiply both sides of an equation by 0, scorch the Earth, and make any progress towards a solution; so don't employ similar anti-strategies on limit problems.

Another issue with your strategy is that in many limit problems, you should try to factor and eliminate any terms from the numerator and denominator when you can, as long as you can confirm that you are not dividing by 0 when doing so. In this case, you could have eliminated (1/x) from the numerator and denominator of the complicated function you arrived at, but you didn't. Sometimes it is correct to multiply the numerator and denominator by the same term, but only when you know that action would ultimately lead to canceling terms later.

Also, when working with limits, you must understand that there are different degrees of approaching 0 and approaching infinity. Simple examples:
$\lim _{x\to \infty }\frac{x}{x^2} = \frac{\infty}{\infty^2} = ??$
$\lim _{x\to \infty }\frac{\frac{1}{x}}{\frac{1}{x^2}} = \frac{0}{0} = ??$
Well, it turns out, in the first example, as you plug-in arbitrarily large values, you get that the denominator is growing much larger than the numerator, and that the entire thing is tending towards 0.
In the second example, as you plug-in arbitrarily large values, you get that the denominator is shrinking a lot faster than the numerator, and that the entire thing is tending towards positive infinity.
This is precisely why, when working with limits, you have to understand that different functions may approach infinity at different rates, or different functions may approach 0 at different rates. You can't treat all zeros the same, and you can't treat all infinities the same. Here's another simple example...
$\lim _{x\to \infty }\frac{2x}{3x} = \frac{2*\infty}{3*\infty} = 2/3$
The numerator is approaching infinity at two-thirds the rate that the denominator is. The value of the limit turns out to be 2/3, which would have been elementary if you just canceled the x term from the numerator and denominator (but, when you do so, you must ensure that you aren't inadvertently dividing by 0, which you are not). But if you fail to eliminate a common term from the numerator and denominator (like you did in your original question), and you plug-in literally infinity for x (like you did in your original question), and you make the mistake of treating all degrees of infinity as being the same (like you did in your original question), then it will seem like the limit is approaching $\frac{\infty}{\infty} = 1??$ This is why you can't use those kinds of strategies when you approach limit problems.

$\endgroup$
0
$\begingroup$

Yes for that limit we have the indeterminate form

$$\lim_{x\to\infty} \frac{\frac{1}{x}}{\sqrt{\frac{1}{x}-\frac{1}{x^2}}}=\frac{0}{0}$$

but we can eliminate the indetermination by

$$\lim_{x\to\infty}\frac{\frac{1}{x}}{\sqrt{\frac{1}{x}-\frac{1}{x^2}}}=\lim_{x\to\infty}\frac{\sqrt{\frac1x} }{\sqrt{\frac1x}}\frac{\sqrt{\frac1x}}{\sqrt{1-\frac{1}{x}}}=\frac{0}1=0$$

therefore it seems there is nothing special to that.

The latter manipuilation explain also the reason for a different result in the second case when we divide numerator and denominator by $\sqrt x$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.