Left and Right-Hand Limits

In some cases, you let x approach the number a from the left or the right, rather than "both sides at once" as usual.

The left- and right-hand limits are the same if and only if the ordinary limit exists. In this case, the left-hand, right-hand, and ordinary limit are equal.


Example.

$$\lim_{x \to 0+} \dfrac{|\sin x|}{\sin x} = 1, \quad\hbox{but}\quad \lim_{x \to 0-} \dfrac{|\sin x|}{\sin x} = -1.$$

Look at the first limit more closely. x approaches 0 from the right. Numbers close to, but to the right of, 0 are small positive numbers: 0.01, for example. Small positive numbers make $\sin x$ positive: $\sin 0.01 \approx 0.01000$ , for example. If $\sin x$ is positive, then $|\sin x| = \sin x$ , so

$$\dfrac{|\sin x|}{\sin x} = \dfrac{\sin x}{\sin x} = 1.$$

(Notice that you don't let x equal 0, so $\sin x \ne 0$ , and the cancellation is legal.)

Therefore,

$$\lim_{x \to 0+} \dfrac{|\sin x|}{\sin x} = \lim_{x \to 0+} 1 = 1.$$

Here's the picture:

$$\hbox{\epsfysize=2in \epsffile{limlr1.eps}}$$

Since the left- and right-hand limits are not the same, $\displaystyle \lim_{x \to 0} \dfrac{|\sin
   x|}{\sin x}$ is undefined.


Example. Suppose

$$f(x) = \cases{2x + 1 & if $x < 1$ \cr 5 & if $x = 1$ \cr 7x^2 - 4 & if $x > 1$ \cr}.$$

Compute $\displaystyle \lim_{x\to 1^+}
   f(x)$ , $\displaystyle \lim_{x\to 1^-} f(x)$ , and $\displaystyle
   \lim_{x\to 1} f(x)$ .

To compute $\displaystyle \lim_{x\to 1^+}
   f(x)$ , I use the part of the definition for f which applies to $x >
   1$ :

$$\lim_{x\to 1^+} f(x) =\lim_{x\to 1^+} (2x + 1) = 3.$$

Likewise, to compute $\displaystyle
   \lim_{x\to 1^-} f(x)$ , I use the part of the definition for f which applies to $x < 1$ :

$$\lim_{x\to 1^-} f(x) =\lim_{x\to 1^-} (7x^2 - 4) = 3.$$

Since the left and right-hand limits are equal, the two-sided limit is defined, and$\displaystyle \lim_{x\to
   1} f(x) = 3$ .

The fact that $f(1) = 5$ does not come into the problem.


Example. Consider the function $f(x)$ whose graph is depicted below:

$$\hbox{\epsfysize=2in \epsffile{limlr2.eps}}$$

Then

$$\lim_{x\to 1^+} f(x) = 1 \quad\hbox{and}\quad \lim_{x\to 1^-} f(x) = 3.$$

Since the left- and right-hand limits are not the same,

$$\lim_{x\to 1} f(x) \hbox{ is undefined.}\quad\halmos$$


Example. Consider the function $f(x)$ whose graph is depicted below:

$$\hbox{\epsfysize=2in \epsffile{limlr3.eps}}$$

Then

$$\lim_{x\to 0^+} f(x) = 1 \quad\hbox{and}\quad \lim_{x\to 0^-} f(x) = 1.$$

Therefore,

$$\lim_{x\to 0} f(x) = 1.$$

The value of $f(0)$ does not affect the existence of the limit. In fact, suppose I change the function as follows:

$$\hbox{\epsfysize=2in \epsffile{limlr4.eps}}$$

Now $f(0)$ is undefined, but

$$\lim_{x\to 0^+} f(x) = 1, \quad \lim_{x\to 0^-} f(x) = 1, \quad \quad\hbox{and}\quad \lim_{x\to 0} f(x) = 1.\quad\halmos$$


Example. Compute $\displaystyle \lim_{x \to 1+} \dfrac{x^2 - 2x - 3}{x - 1}$ .

Plugging in gives $\dfrac{-4}{0}$ . The limit is undefined. But I can say more.

Try plugging in a number close to 1: When $x = 1.001$ ,

$$\dfrac{x^2 - 2x - 3}{x - 1} \approx -4000.$$

It looks as though $\dfrac{x^2 - 2x -
   3}{x - 1}$ is getting big and negative. In fact,

$$\lim_{x \to 1+} \dfrac{x^2 - 2x - 3}{x - 1} = -\infty.$$

To why this is true, remember that x is approaching 1 from the right. This means that $x - 1$ will be small and positive. On the other hand, $x^2 - 2x - 3 \to -4$ . Since the top is negative and the bottom is positive, the result must be negative.

As far as size goes, I have

$$\dfrac{\hbox{nonzero number}}{\hbox{small number}} = \hbox{big number}.$$

Since the result should be big and negative, it is reasonable that it is $-\infty$ .

Another way to see this is to draw the graph near $x = 1$ . As you move toward 1 from the right, the graph goes downward toward $-\infty$ .

$$\hbox{\epsfysize=1.8in \epsffile{limlr5.eps}}\quad\halmos$$


I noted earlier that if

$$\dfrac{f(x)}{g(x)} \to \dfrac{\hbox{nonzero number}}{0},$$

then the two-sided limit $\displaystyle
   \lim_{x\to c} \dfrac{f(x)}{g(x)}$ is undefined. As the example above shows, the situation is different with one-sided limits.

If, in this situation, $g(x)$ has the same sign for all x's sufficiently close to c and greater than c, then the right-hand limit $\displaystyle \lim_{x\to c^+}
   \dfrac{f(x)}{g(x)}$ will be either $+\infty$ or $-\infty$ . The specific sign depends on the signs of the top and the bottom of the fraction.

Likewise, if $g(x)$ has the same sign for all x's sufficiently close to c and less than c, then the left-hand limit $\displaystyle \lim_{x\to c^-}
   \dfrac{f(x)}{g(x)}$ will be either $+\infty$ or $-\infty$ . Again, the specific sign depends on the signs of the top and the bottom of the fraction.

The "same-sign" condition will be satisfied, for example, if f and g are polynomials --- that is, if $\dfrac{f(x)}{g(x)}$ is a rational function. It will also be satisfied by functions like

$$\dfrac{x - 3}{x^{1/3} - 2} \quad\hbox{as}\quad x\to 8.$$


Example. Compute $\displaystyle \lim_{x\to -3^+} \dfrac{x + 1}{x + 3}$ .

Plugging $x = -3$ in gives $\dfrac{-2}{0}$ . Since $\dfrac{x + 1}{x + 3}$ is a rational function, the right-hand limit $\displaystyle \lim_{x\to -3^+} \dfrac{x + 1}{x + 3}$ is either $+\infty$ or $-\infty$ ; I have to determine which of the two it is. I'll look at the top and the bottom separately.

As $x\to -3^+$ , $x + 1 \to -2$ .

As for the bottom, since x is approaching -3 from the right, I'm considering x's greater than -3. Thus, $x > -3$ , so $x + 3 > 0$ --- $x + 3$ is positive.

Since $x + 1$ is approaching a negative number and $x + 3$ is approaching a positive number, the quotient is negative. Therefore,

$$\lim_{x\to -3^+} \dfrac{x + 1}{x + 3} = -\infty.$$

I can also see this if I take a number close to -3 but to the right of -3 --- $x = -2.99$ , for example --- and plug it in:

$$\dfrac{x + 1}{x + 3} = \dfrac{-2.99 + 1}{-2.99 + 3} = \dfrac{-1.99}{0.01} = -199.$$

I got a large negative number, which suggests that the limit should be $-\infty$ .

I could also see this by graphing the function, as in the previous example.

You might ask: "Which of these methods is the best?" I feel that for a first course in calculus, all three are acceptable. (But some people might disagree, so you should be careful to ask.)

However, while plugging in numbers and drawing graphs provide support for a conclusion, they don't really provide a proof. Graphs can be deceiving. And when you plug in a number, how do you know that the number you chose is "typical"? The first method --- reasoning about signs using inequalities --- is much closer to a rigorous proof of the result.


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