Limits

Calculus I. Also known as: math 151, math 2413, math 1550, math 1910, math 140, calc 1, calc i, calculus 1, calculus i, differential calculus, calculus for science and engineering, math 121

A limit describes what a function approaches near a point. Learn to estimate limits, use one-sided limits and evaluate them with algebra.

What a limit says

The limit of f(x) as x approaches a is the value that f(x) gets closer and closer to when x is near a but not equal to a. We write lim(x -> a) f(x) = L. The function does not even need to be defined at a. A limit studies the neighbourhood of a, never the point itself, so f(a) and the limit can be completely different, or f(a) can be missing.

Reading a limit from numbers

Take g(x) = (x^2 - 16)/(x - 4). At x = 4 it gives 0/0, which is undefined. Now try nearby inputs: g(3.9) = 7.9, g(3.99) = 7.99 and g(3.999) = 7.999 from the left, and g(4.001) = 8.001 from the right. Both sides head toward 8, so the limit is 8. A table suggests the answer; algebra proves it.

One-sided limits

The left-hand limit uses only x < a and the right-hand limit uses only x > a. The two-sided limit exists exactly when both one-sided limits exist and are equal. If the graph jumps, or if either side grows without bound, the two-sided limit does not exist. Never average two different one-sided values.

An algebra toolkit

  1. Try direct substitution first. A finite number means you are done.
  2. If you get 0/0, factor and cancel. Here x^2 - 16 = (x - 4)(x + 4), so for x not equal to 4 the fraction is just x + 4, and the limit is 8.
  3. If square roots cause the 0/0, multiply top and bottom by the conjugate.
  4. If you get a nonzero number over 0, expect the values to grow without bound, and check the sign on each side.
  5. Use the standard result that sin(x)/x approaches 1 as x approaches 0 (x in radians). By rescaling, sin(kx)/x approaches k.

Worked example

Find the limit as x approaches 9 of (sqrt(x) - 3)/(x - 9).

Substitution gives 0/0. Multiply top and bottom by the conjugate sqrt(x) + 3. The top becomes (sqrt(x))^2 - 9 = x - 9, which cancels the bottom, leaving 1/(sqrt(x) + 3) for x not equal to 9. As x approaches 9 this tends to 1/(3 + 3) = 1/6. A numeric check agrees: at x = 9.0001 the original fraction is about 0.1667.

Unbounded behaviour

As x approaches 0, the value of 1/x^2 grows without bound because the denominator is tiny and positive on both sides. We write this as +infinity to describe the behaviour, although strictly the limit does not exist as a real number.

Common mistakes

  • Stopping at 0/0 and declaring the limit is 0 or does not exist. It is only a signal to do more algebra.
  • Mixing up f(a) with the limit.
  • Cancelling terms instead of factors, such as crossing out the x in (x + 4)/x.
  • Forgetting that sin(x)/x -> 1 needs radians.

Practice questions

  1. What is the limit as x approaches 3 of (x^2 - 9)/(x - 3)?

  2. What is the limit as x approaches 2 of (x^2 + x - 6)/(x - 2)?

  3. What is the limit as x approaches 4 of (sqrt(x) - 2)/(x - 4)?

  4. What is the limit as x approaches 0 of sin(5x)/x, with x in radians?

  5. Let f(x) = 2x - 1 for x < 3 and f(x) = x^2 - 2 for x >= 3. What is the limit of f(x) as x approaches 3 from the right?

  6. Let f(x) = 2x - 1 for x < 3 and f(x) = x^2 - 2 for x >= 3. What is the two-sided limit of f(x) as x approaches 3?