Algebraic Evaluation of Limits

Calculus Tutorial

Algebraic Evaluation of Limits


We are going to use a simple example to explain key concepts in limit.

Sample Problem

Evaluate lim x->1 (x^2 – 1)/(x – 1)


lim x->1 (x^2 – 1)/(x – 1)

= lim x->1 (x – 1)(x + 1)/(x – 1)

= lim x->1 (x + 1) [1. Why we can cancel x – 1?]

= 1 + 1 [2. Why we can substitute x = 1 into x + 1?]

= 2

The above procedures are widely discussed in textbooks. To understand why we can do that, we need to go back to basic concepts about limit of a function.

1. When x = 1, (x – 1) is 0. Why we can cancel this term? Isn’t it 0/0 which is undefined?

By definition, limit of a function is the behaviour of a function when we are approaching to a particular point. It does not mean that we evaluate the function at that particular point. In the above example, we want to evaluate the function close to x = 1, but not at x = 1. As long as x is close to 1 (but not equal to 1), the term x – 1 is not equal to 0. Therefore, we can cancel this term in numerator and in denominator.

2. Why do we simply substitute x = 1 in x + 1 to evaluate the limit?

When a function f(x) is continuous at, say x = c, lim x->c f(x) = f(c). If a function is continuous at x = c, whether we approach x = c from the left side or from the right side, we will get to the same point (c, f(c)). In the above example, as the function x + 1 is continuous at x = 1, we can simply substitute x = 1 to evaluate the limit.

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