In calculus, limits are fundamental in understanding the behavior of functions as they approach specific points. Proofs involving limits help establish the validity of various mathematical statements and theorems. They are essential for confirming properties of functions, continuity, and differentiability. Here, we present three diverse and practical examples that illustrate different methods of proving limits.
Polynomial functions are continuous everywhere, but proving their limits can reinforce our understanding of how limits work in general.
When evaluating
[ \lim_{x \to 3} (2x^2 + 5x - 4) ]
we will demonstrate that we can find this limit directly by substitution.
To prove this, we substitute 3 directly into the polynomial:
[ \lim_{x \to 3} (2x^2 + 5x - 4) = 2(3^2) + 5(3) - 4 ]
[ = 2(9) + 15 - 4 ]
[ = 18 + 15 - 4 ]
[ = 29. ]
Thus, we conclude that
[ \lim_{x \to 3} (2x^2 + 5x - 4) = 29. ]
Rational functions can have limits that require simplification before evaluation. This example will show how to evaluate the limit of a rational function as it approaches a point where it may be indeterminate.
Consider the limit:
[ \lim_{x \to 1} \frac{x^2 - 1}{x - 1}. ]
First, we notice that substituting 1 directly gives us the indeterminate form ( \frac{0}{0} ). Thus, we will simplify the expression:
[ \frac{x^2 - 1}{x - 1} = \frac{(x - 1)(x + 1)}{x - 1}. ]
For ( x \neq 1 ), we can cancel the ( (x - 1) ) terms:
[ = x + 1. ]
Now, we can evaluate the limit:
[ \lim_{x \to 1} (x + 1) = 1 + 1 = 2. ]
Thus,
[ \lim_{x \to 1} \frac{x^2 - 1}{x - 1} = 2. ]
The epsilon-delta definition of a limit is a rigorous way to prove limits. This method is particularly useful in formal proofs and higher mathematics.
Let’s prove that:
[ \lim_{x \to 2} (3x - 1) = 5. ]
We want to show that for every ( \epsilon > 0 ), there exists a ( \delta > 0 ) such that if ( 0 < |x - 2| < \delta ), then ( |(3x - 1) - 5| < \epsilon. )
Starting from the expression, we have:
[ |(3x - 1) - 5| = |3x - 6| = 3|x - 2|. ]
We need:
[ 3|x - 2| < \epsilon \Rightarrow |x - 2| < \frac{\epsilon}{3}. ]
Thus, we can choose ( \delta = \frac{\epsilon}{3} ). So, if ( 0 < |x - 2| < \delta ), we have:
[ |(3x - 1) - 5| < \epsilon. ]
This confirms that
[ \lim_{x \to 2} (3x - 1) = 5. ]