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# Limits

Let y be a function of x. We ask what is the number that y approaches as x approaches a given number.

Consider the function

$\begin{array}{rcl}y(x) &= &4,\ 0 \leq x < 2 \\ &= &3,\ x = 2 \\ &= &4,\ 2 < x \leq 8\end{array}$

The function is shown below.

What number does y approach as x approaches 2? We use an arrow to replace the word approaches. We also say tends to in place of approaches.

It is clear that y $$\to$$ 4 as x $$\to$$ 2 from the left and y $$\to$$ 4 as x $$\to$$ 2 from the right. Note that the value of the function itself, y(2) = 3 is different from the value that y approaches.

We write this as follows:

$\lim_{x\to2} y = 4$

Consider the function

$y = \frac{x^2 - 4}{x - 2}$

What is the limit of as x tends to 2?

If we put x = 2, we get y = 0/0, which is indeterminate.

Note that the question is not what is the value of y when x = 2. (At this value of x the function is not defined.)

The question is, what value does y approach as x approaches 2? That is, we let x get closer and closer to 2, but not equal to 2. To answer this we write y as,

$\begin{array}{rcl}y(x) &= &\frac{x^2 - 4}{x - 2} \\ &= &\frac{(x - 2)(x + 2)}{x - 2} \\ &= &x + 2\end{array}$

We have divided the top and bottom by (x – 2). Now dividing by a number is permitted as long as it is not equal to zero. So the above maneuver is valid as long as x is not equal to 2. But that is all right for our problem, since we are not letting x = 2; we are only letting x approach 2.

x can get as close to 2 as we like, but we do not equate it to 2.

Thus, y(x) = (x + 2) as long as x is not equal to 2. Now we can see that y approaches 4 as x approaches 2.

$\lim_{x\to2} y = 4$

1. What is the limit of

$\frac{x^2 – 4x + 4}{x - 2}$

as x$$\to$$2?

$\frac{x^2 – 4x + 4}{x - 2}\ = \frac{(x - 2)^2}{x - 2} = x - 2$

Therefore,

$\lim_{x\to2} \frac{x^2 – 4x + 4}{x - 2}\ = \lim_{x\to2}\ (x - 2) = 0$

2. What is the limit of 1/x as x tends to 0?

If x is positive, we see that as x becomes smaller and smaller 1/x becomes larger and larger. No matter how large a number you might pick, we can overtake that by making x close enough to zero. So we might say that the limit is ∞.

But if x approaches 0 from the negative side, we see that the limit is -∞. One of the important properties of a limit is that it should be unique and independent of how we approach the value of x. Therefore, in this case, we can say that there is no limit.

3. Evaluate

$\lim_{x\to\infty} \frac{7x^8 + 6x^5 - 3x^2 - 8}{4x^8 + 3x^3}$

The trick is to first divide the numerator and denominator by x8. You can then take the limit.

7/4

4. Find

$\lim_{x\to2} \frac{\sqrt{x} - \sqrt{2}}{x^2 - 4}$
$\frac{\sqrt{x} - \sqrt{2}}{x^2 - 4} \\ = \frac{\sqrt{x} - \sqrt{2}}{(x - 2)(x + 2)} \\ = \frac{\sqrt{x} - \sqrt{2}}{(\sqrt{x} - \sqrt2)(\sqrt{x} + \sqrt2)(x + 2)} \\ = \frac{1}{(\sqrt{x} + \sqrt2)(x + 2)}$

Therefore,

$\lim_{x\to2} \frac{\sqrt{x} - \sqrt{2}}{x^2 - 4} = \lim_{x\to2} \frac{1}{(\sqrt{x} + \sqrt2)(x + 2)} \\ = \frac{1}{8\sqrt2}$

5. Evaluate

$\lim_{h\to0} \frac{(x + h)^2 - x^2}{h}$
$\frac{(x + h)^2 - x^2}{h}\ = \frac{(x^2 + 2xh + h^2) - x^2}{h}\ \\ = \frac{2xh + h^2}{h}\ = 2x + h$
$\lim_{h\to0} \frac{(x + h)^2 - x^2}{h}\ = \lim_{h\to0}\ (2x + h) = 2x$