How to evaluate the limit of a real-valued function

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In mathematics, the definition of the limit of a function does not cover how to evaluate the limit of a real-valued function: it is not computational, on the face of it.

1 Case of continuous functions
2 Limits not directly accessible
3 Limits do not always exist
4 L'Hôpital's rule, and division by zero
5 A few notable limits
6 The "squeeze theorem"
7 Approaching a limit
8 See also

[edit] Case of continuous functions

For "well-behaved" functions (i.e., continuous ones), the limit as x approaches c can be found by directly substituting c for x. For example, if f(x) = 7 as x approaches 32, the limit is 7 (the limit of a constant is a constant). Another example is f(x) = 2x - 5; in that situation, as x approaches 3, f(x) approaches f(3) = 2·3 - 5 = 1.

This is only a tool for this special case. The definition of limit doesn't involve the point f(c) at all. Its value can be arbitrary, or can even not exist, and the limit won't change. But if the function is continuous at that point, the limit and the function value at that point happen to be the same.

[edit] Limits not directly accessible

Limits are more interesting when they are unreachable; for example: if f(x) = (x³ − 1) / (x − 1), then x cannot be set equal to 1 (as that would result in division by zero). More formally, the value 1 is not in the function domain.

Here, however, f(x) does approach some number c as x approaches 1.

As x approaches 1, then f(x) approaches 3. Note, however, that f(x) is undefined at x=1

x	f(x)
0.9	2.71
0.99	2.9701
0.999	2.997001
1	(undefined)
1.001	3.003001
1.01	3.0301
1.1	3.31

We see that, as x approaches 1, f(x) approaches 3; however, x is never equal to 1, and f(x) never equals 3.

The limit can be verified using algebra. Note that (x² + x + 1)(x − 1) = (x³ − 1), and define g(x) = x² + x + 1, so that g(x) = f(x) (for any x ≠ 1). That is, g(x) and f(x) are identical at every point except at x = 1; f(x) has a 'hole' in its domain at 1. Via direct substitution (x = 1) one can determine the limit of g(x) to be 1² + 1 + 1 = 3, which is the value f(x) approaches as it approaches the hole.

Here we have used the following rule: if it so happens that if g(x) = f(x) for all values of x, except for c; then, so long as g(x) has a limit, that limit will be equal to the limit of f(x).

The above also shows that whether f(c) exists has no bearing on whether the limit of f(x) (as x approaches c) exists. Should f(c) exist, its value has no bearing on the limit.

[edit] Limits do not always exist

Not every function has a limit at every point. Consider:

As x approaches 0, f(x) = |x| / x does not approach a limit; the value of f(x) is -1 if x<0 and +1 if x>0. No number L can serve as a limit; we say that the "one-sided limit from the left as x approaches 0" is -1 and the "one-sided limit from the right" is +1.
As x approaches 0, f(x) = 1 / x does not approach a limit; for x>0 it is unbounded above and approaches +∞, for x<0 it is unbounded below and approaches -∞.
As x approaches 0, f(x) = sin (1/x) does not approach a limit, as it oscillates faster and faster.

[edit] L'Hôpital's rule, and division by zero

When trying to evaluate a limit by simply substituting c into the function, one often would have to divide by zero, which is, of course, impossible. Here l'Hôpital's rule helps: If f(c)->0, g(c)->0, and both derivatives f'(c) and g'(c) are defined, then:

$\lim_{x\to c}{f(x)\over g(x)}=\lim_{x\to c}{f'(x)\over g'(x)}$

[edit] A few notable limits

$\lim_{x \to 0}\frac{\sin x}{x} = 1$

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

$\lim_{x \to 0}\frac{e^x-1}{x} = 1$

$\lim_{x \to 0}\frac{a^x-1}{x} = \ln (a)$

$\lim_{x \to 0}\frac{\ln (1+x)}{x} = 1$

$\lim_{x \to 0}\frac{\log_a (1+x)}{x} = \frac{1}{\ln (a)}$

$\lim_{n\to\infty} \left( 1 + \frac{a}{n} \right)^n = e^a$

$\lim_{n\to\infty} \left( 1 - \frac{1}{n} \right)^n = \frac{1}{e}$

$\lim_{n\to\infty} \sqrt[n]{n} = 1$

[edit] The "squeeze theorem"

The "squeeze theorem" (or "pinching theorem," as it is also known) is a theorem telling us that if three functions f(x), g(x) and h(x) are given such that h(x) ≤ f(x) ≤ g(x) and if the limit, L, (as x approaches c) of h(x) is equal to the limit (as x approaches c) of g(x), then the limit of f(x) not only exists, but is also equal to L. The function f(x) is "squeezed" between g(x) and h(x).

[edit] Approaching a limit

One might ask whether there is some relationship between |f(x) − L| (the absolute value [of the function, at x, minus its limit, L, as x approaches c]) and |x − c| (the absolute value [of x minus the number it is approaching]). For every number, ε > 0, there is some number, δ > 0; such that, if 0 < |x − c| < δ, then |f(x) − L| < ε. In other words, if the distance between x and c is less than δ, then the distance between f(x) and L is less than ε. The ε chosen may be made arbitrarily small. The above statement is one common definition of the limit of a real-valued function.

For example; the limit of (3x − 2), as x approaches 3, is 7; f(3) = 7. If one determines that the absolute value [of the function, at x, minus its limit (as x approaches 3)], should be less than 0.003 (that is, one is attempting to determine what value of x will generate an f(x) within 0.003 of the limit of f(x), as x approaches c = 3); then, one can write: |(3x − 2) - 7| = |3x − 9| = 3|x − 3| < 0.003. Noting that 0 < |x − c| < δ and that, in this situation, c = 3; we can write 0 < |x − 3| < δ and since 3|x − 3| < 0.003; it is only logical to conclude that |x − 3| < 0.003 / 3 = 0.001; and thus, any value within 0.001, of 3, will produce a value within 0.003 of f(x)'s limit (as x approaches 3); that is, a value within 0.003 of 7. For instance, f(3.001) = 7.003.