On one of my past calculus tests, I struggled with a multi-part question that asked to "Evaluate the limits that exist, or prove that they don't", as I would often be plagued by doubt wondering if each limit exists, and wound up wasting time changing my answers.

Now with my final upcoming later this week, the last thing I want to do is spend any time second-guessing myself as to whether or not the limit exists. What sort of techniques can I use to intuitively determine the existence of a limit of a function before delving into evaluation or a proof of non-existence?

I'm in a first year single variable Calculus course; here are a few limits.

$$\lim_{x \to 0} \frac{\tan(3x^2) + \sin^2(5x)}{x^2}$$

$$\lim_{x \to 3^+} \frac{\sqrt{x - 3}}{|x - 3|}$$

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

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    $\begingroup$ (Accurately) Plotting the function in question helps a great deal. $\endgroup$ – J. M. is a poor mathematician Apr 12 '11 at 11:48
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    $\begingroup$ If we restrict our attention to a limit of a multivariable function, it is sometimes helpful to taking the limit along a particular path, especially when you try to prove that it does not exist. If there are two ways of approaching that yield different answers, then the original limit diverges. For example, let's consider $\lim_{(x, y) \to (0, 0)} \frac{x^2 - y^2}{x^2 + y^2}$. Taking the limit along the line $y = mx$, we obtain $\frac{1 - m^2}{1 + m^2}$. For different choices of $m$, we clearly obtain different results, so we can deduce that the limit does not exist. $\endgroup$ – Sangchul Lee Apr 12 '11 at 13:16
  • $\begingroup$ Your question is actually rather broad -- e.g. I can't be sure whether the above comments are appropriate for you or not. What kind of calculus class are you talking about: first semester, second semester, multivariate...? Also do you mean limits of functions, sequences, series...? Giving a few examples wouldn't hurt. $\endgroup$ – Pete L. Clark Apr 12 '11 at 15:20
  • $\begingroup$ The title mentions functions but I realized that I omitted that from the body, whoops. I'll clarify with examples. $\endgroup$ – Mana Apr 12 '11 at 15:25
  • $\begingroup$ @Mana: Instead of $\large...$, use $$...$$; that will scale things to displaystyle, which also spaces out the fractions and puts the limits under the $\lim$ operator. $\endgroup$ – Arturo Magidin Apr 12 '11 at 15:41

You want to be on the lookout for familiar limits, relating new limits to limits you have experience with.

The first limit can be manipulated so that it is in terms of famous limits: notice that $$\frac{\tan(3x^2)}{x^2} = 3\left(\frac{\tan(3x^2)}{3x^2}\right) = 3\left(\frac{\sin(3x^2)}{3x^2}\right)\left(\frac{1}{\cos(3x^2)}\right)$$ and $$\frac{\sin^2(5x)}{x^2} = \left(\frac{\sin(5x)}{x}\right)\left(\frac{\sin(5x)}{x}\right) = 25\left(\frac{\sin(5x)}{5x}\right)\left(\frac{\sin(5x)}{5x}\right).$$ When $x\to 0$, the cosine gives no trouble, and the limits involving the sine are of the form $$\lim_{u\to 0}\frac{\sin(u)}{u}$$ which you should recognize. This suggests that the limit exists, and how to compute it.

The second limit is obfuscated: since $x\to 3^+$, you can drop the absolute value sign (since $x\gt 3$ so $|x-3| = x-3$) and then you have $$\frac{\sqrt{x-3}}{|x-3|} = \frac{\sqrt{x-3}}{x-3}$$ which is of the form $\frac{\sqrt{u}}{u}$. It should now be easy to see that as $u\to 0^+$, this limit does not exist (simplify it).

For the third limit, you have a polynomial in $\sqrt{x}$ divided by a polynomial in $\sqrt{x}$, and both go to $0$ when $x\to 1$. You should expect to be able to factor out $(\sqrt{x}-1)$ from both numerator and denominator and cancel them. Since the denominator is $(\sqrt{x}-1)(\sqrt{x}+1)$, there will be only a single factor of $\sqrt{x}-1$, so the limit should exist after you cancel those factors (the new denominator will not be $0$ at $x=1$). Indeed: the numerator is $u^4 - u = u(u^3-1)=u(u-1)(u^2+u+1)$ (with $u=\sqrt{x}$), the denominator is $(u-1)(u+1)$, so you can cancel and evaluate.

  • $\begingroup$ I guess you meant "It should now be easy to see...". $\endgroup$ – Hans Lundmark Apr 13 '11 at 7:02
  • $\begingroup$ @Hans: Heh; yes, it should. $\endgroup$ – Arturo Magidin Apr 13 '11 at 15:35

Intuition comes with experience, and is hard to capture by simple rules. Here's a more practical advice: Start rewriting the expression just as you would when computing the limit, and then you will usually notice after a while if it works out or if there is some kind of obstacle that causes the limit not to exist.


if you remember very early on you show that $\sin(small) = small + \ldots, \cos(small) = 1 - \frac{1}{2}small^2 + \ldots$ and $\tan(small) = small + \ldots$ also the binomial theorem $(BIG + small)^n = BIG^n + nBIG^{n-1}small + \ldots$.

how do you translate these in finding limits. take your first question $\lim_{x \to 0} \frac{\tan(3x^2) + \sin^2(5x)}{x^2}?$ this reduces to finding $\lim_{x \to 0}\frac{3x^2 +(5x)^2 }{x^2}$ which is 28. whenyou have small arguments for $\sin$ or $\tan$ you can drop the $\sin$ or $\tan$ and the quotient you have to now deal with is a rational function which should be easier.

finding limits when $x$ aproaches a non zero can be handled with a change of variable.


All these limits I would start off just by estimating the dominating power in the numerator/denominator. My thought process upon seeing them went like this:

First limit: "$\sin$ and $\tan$ look like $x$ near zero, they both get squared (inside or outside the function doesn't matter), so they look like $x^2$. $x^2$ in the denominator ties with them, so we should look at coefficients to figure out what the limiting ratio will be. Probably $\frac{3+5}{1}$ = 8, since coefficients tend to move in and out of $\sin$ and $\tan$ in the limit."

Second limit "Approaches 3 like a square root in the numerator and like $x$ in the denominator, higher power wins out so it goes to infinity." (If we need to know whether it's positive or negative infinity, we check the signs to find that everything is positive no matter what $x$ is so it's $+\infty$.)

Third limit: "Factor out a $\sqrt{x}$ in the numerator, which then doesn't matter since it's going to 1. Now the top approaches like $x^{\frac{3}{2}}$ and the bottom approaches like $x$, higher power wins, thus it's going to zero."

This sort of reasoning doesn't give proofs and wouldn't be accepted as an answer. Indeed, I guessed the wrong limit for the first one, because I forgot to square the coefficients in the $\sin$ function. But figuring out which pieces of the equation "win" (and remembering things like $\sin$ and $\tan$ looking like $x$, or $\cos$ looking like $1-x^2$, or lower power < higher power < exponential < factorial) can really give you a very accurate guess as to what limit you should expect.


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