Limits to memorize

You should memorize the following limits. If you fail to memorize the last three, you’ll waste a lot of time trying to figure them out.

• 

  ( is a horizontal line, so the limit — which is the function height — must equal c regardless of the x-number.)

• 
• 
• 
• 
• 
• 
• 

Plug-and-chug limits

Plug-and-chug problems make up the second category of easy limits. Just plug the x-number into the limit function, and if the computation results in a number, that’s your answer. For example,

This method works for limits involving continuous functions and functions that are continuous over their entire domains. These are well-duh limit problems, and, to be perfectly frank, there’s really no point to them. The limit is simply the function value.

The plug-and-chug method works for any type of function unless there’s a discontinuity at the x-number you plug in. In that case, the number you get after plugging in is not the limit. And if you plug the x-number into a limit like and you get any number (other than zero) divided by zero — like — then you know that the limit does not exist.

Factoring

Here’s an example. Evaluate .

1. Try plugging 5 into x — you should always try substitution first.

   You get — no good, on to plan B.

2. can be factored, so do it.
   
3. Cancel the from the numerator and denominator.
   
4. Now substitution will work.
   

The function you got after canceling the , namely , is identical to the original function,, except that the hole in the original function at (5, 10) has been plugged. Note that the limit as x approaches 5 is 10, which is the height of the hole at (5, 10).

Conjugate multiplication

Try this method for fraction functions that contain square roots. Conjugate multiplication rationalizes the numerator or denominator of a fraction, which means getting rid of square roots. Try this one: Evaluate .

1. Try substitution.

   Plug in 4: That gives you — time for plan B.

2. Multiply the numerator and denominator by the conjugate of , which is .

   Now do the rationalizing.

   
3. Cancel the (x – 4) from the numerator and denominator.
   
4. Now substitution works.
   

As with the factoring example, this rationalizing process plugged the hole in the original function. In this example, 4 is the x-number, is the answer, and the function has a hole at .

Miscellaneous algebra

When factoring and conjugate multiplication don’t work, try other basic algebra like adding or subtracting fractions, multiplying or dividing fractions, canceling, or some other form of simplification.

Evaluate .

1. Try substitution.

   Plug in 0: That gives you — no good.

2. Simplify the complex fraction (that’s a big fraction that contains little fractions) by multiplying the numerator and denominator by the least common denominator of the little fractions, namely .

   Note: Adding or subtracting the little fractions in the numerator or denominator also works in this type of problem, but it’s a bit longer than the method here.

   
3. Now substitution works.
   

Horizontal asymptotes

Horizontal asymptotes and limits at infinity go hand in hand — you can’t have one without the other. For a rational function like , determining the limit at infinity or negative infinity is the same as finding the location of the horizontal asymptote.

Here’s what you do. First, note the degree of the numerator (that’s the highest power of x in the numerator) and the degree of the denominator. You’ve got three cases:

• If the degree of the numerator is greater than the degree of the denominator, for example , there’s no horizontal asymptote and the limit of the function as x approaches infinity (or negative infinity) does not exist.
• If the degree of the denominator is greater than the degree of the numerator, for example , the x-axis (the line ) is the horizontal asymptote and .
• If the degrees of the numerator and denominator are equal, take the coefficient of the highest power of x in the numerator and divide it by the coefficient of the highest power of x in the denominator. That quotient gives you the answer to the limit problem and the height of the asymptote. For example, if , and h has a horizontal asymptote at .

Substitution doesn’t work for the problems in this section. If you try plugging into x in any of the rational functions in this section, you get , but that does not equal 1. A result of tells you nothing about the answer to a limit problem.

Solving limits at infinity

Let’s try some algebra for the problem .

1. Try substitution — always a good idea.

   No good. You get , which does not equal zero and which tells you nothing (see the related Warning in the previous section). On to plan B.

   Because contains a square root, the conjugate multiplication method seems like a natural choice, except that that method is used for fraction functions. Well, just put over the number 1 and, voilà, you’ve got a fraction: . Now do the conjugate multiplication.

2. Multiply the numerator and denominator by the conjugate of and simplify.
   
3. Now substitution does work.