Direct Substitution Is The First Thing You Try
Plug the value into the function. If you get a number, you're done. That's it. I see students skip straight to factoring or rationalizing because they think limits are supposed to be complicated. They're not always. The moment your answer has a zero in the denominator after plugging in, that's when you know you have something that needs more work. When direct substitution gives you 0/0 or infinity/infinity, you've hit an indeterminate form. Those aren't answers—they're signs you need a different technique. Factoring is the first tool most people reach for, and for polynomial expressions it works about eighty percent of the time. Take (x^2 - 4)/(x - 2) as x approaches 2. Factor the top to (x-2)(x+2), cancel the (x-2) terms, and you're left with x+2, which equals 4 at x=2. Straightforward. Rationalizing shows up when you have square roots involved. Multiply by the conjugate, simplify, and usually the problematic term cancels out. Trigonometric limits often require either a standard limit identity or a L'Hôpital's Rule application. I used to memorize all the trig limit identities cold, but honestly I just reference them when needed now. The two main ones are lim(x->0) sin(x)/x = 1 and lim(x->0) (1-cos(x))/x = 0. Everything else builds from those.
L'Hôpital's Rule itself is elegant but easily misapplied. The rule says if you have an indeterminate form, you can take the derivative of the top and the derivative of the bottom separately and evaluate the limit again. But it only works for 0/0 and infinity/infinity forms. If direct substitution gives you something like 5/0, that's not indeterminate—that's either undefined or a vertical asymptote, and L'Hôpital's won't help you there. I've seen people apply it to 5/0 and wonder why they get the wrong answer. It happens constantly in office hours. Here's a case that always catches people off guard. A student once brought me the limit as x approaches 0 of (e^(-1/x^2))/x. Direct substitution gives you 0/0 on the surface, but this one needs you to recognize that e^(-1/x^2) decays faster than any polynomial grows near zero. The trick is to substitute u = 1/x and then use L'Hôpital's repeatedly. I had to work through three applications of L'Hôpital's before it resolved. That function is smooth everywhere else and its limit is actually 0, which is counter-intuitive because the function looks wildly unstable near the origin when you graph it. The one situation where none of these techniques work cleanly is when you have a piecewise function with a jump discontinuity. If f(x) equals x+1 for x less than 2 and equals x-1 for x greater than or equal to 2, then the limit as x approaches 2 doesn't exist because the left-hand limit is 3 and the right-hand limit is 1. You have to check both sides separately. Many textbooks gloss over this, but it comes up on exams frequently enough that you should know to verify both directions before declaring a limit exists.
When Numbers Get Large
Limit at infinity problems follow similar logic but require you to look at the dominant terms. For rational functions, the degree of the numerator and denominator determines the behavior. If the numerator's degree is higher, the limit diverges. If the denominator's degree is higher, the limit is zero. If they're equal, the limit is the ratio of the leading coefficients. This shortcut saves you from doing algebra you don't need to do. Square root expressions at infinity require extra care. The expression sqrt(x^2 + 3x) - x as x approaches infinity looks like infinity minus infinity, which is another indeterminate form. Multiply by the conjugate sqrt(x^2+3x)+x over itself, simplify, and you find the limit is 3/2. Without the conjugate step, you'd be stuck guessing. One limitation worth noting: computational tools like Wolfram Alpha or a TI-89 will give you the right answer for most standard limit problems, but they don't teach you the process. If you're in a calculus course, you need to know the steps manually. The tool output won't earn you credit, and more importantly, understanding the mechanics is what lets you handle problems the software wasn't trained on. I've had colleagues who rely entirely on symbolic computation and still struggle when a problem has a non-standard form that trips up the algorithm.
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