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# Applications of One-Sided Limits

## Determine if a limit exists and, if so, its value.

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Practice Applications of One-Sided Limits
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Introduction to Limits

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### Definition of a Limit

Definition:

The notation   $\lim_{x \rightarrow x_0} f(x) = L$    means that as x approaches (or gets very close to) x, the limit of the function f ( x ) gets very close to the value L

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Basically, a limit is the value a function approaches at a certain point. Limits can be found by:

• plugging the x-value into the equation
• looking at a graph and estimating the y-value for a function at that point
• plugging the equation into a calculator and using a table to see what value the function approaches from the left and right sides
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Write using limit notation:

1. Write the limit of $7x^3 + \sqrt{2x} + 3x - 5$ as $x$ approaches $a$ from the left.
2. Write the limit of $f(m)$ as $m$ approaches $a$.
3. Write the limit of $g(z)$ as $z$ approaches $b$.

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Find the following limits at x = 0:

1.

x -0.2 -0.1 -0.01 0 0.01 0.1 0.2
) 2.993347 2.998334 2.999983 Undefined 2.999983 2.998334

2.993347

2.

x -0.2 -0.1 -0.01 0 0.01 0.1 0.2
) 0.993347 0.998334 0.999983 Undefined 1.000001 1.000012 1.000027

3.

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Find the following limits:

1. $\lim_{x \rightarrow 0} \frac{5x} {2}$
2. $\lim_{x \rightarrow 4} \sqrt{x}$
3. $\lim_{x \rightarrow 0} \frac{sin x} {x}$
4. $\lim_{x \rightarrow 0} \frac{3x} {\sqrt{x + 1} - 1}$
5. $\lim_{x \rightarrow 0} \frac{1-cos x} {x^2}$

### One-Sided Limits

If the value that the function approaches differs on the left and the right, you can use one-sided limits to determine the value.

What is the limit of this function as x approaches 0 from the left? From the right?

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Limits from the left are written with a - after the number, from the right has a +.

Tip: The sign corresponds to the sides of the y-axis. The right side is positive, the left is negative.

#### Does the Limit Exist?

For a limit to exist, the limit from the right side must be equal to the limit from the left. If the right-hand limit does not equal the limit from the left then the limit does not existFor example, in the graph above, $\lim_{x\to0^-} \ne \lim_{x\to0^+}$. Therefore  $\lim_{x\to0}$ does not exist.

To determine if the limit of a piecewise function (a function with two or more parts) exists, you must see if the right-hand and left-hand limits are equal.

Remember that we are not concerned about finding the value of ) at but rather near . So, for < 1 (limit from the left),

$\lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1^{-}} (3 - x) = (3 - 1) = 2$

and for > 1 (limit from the right),

$\lim_{x \rightarrow 1^+} f(x) = \lim_{x \rightarrow 1^+} (3x - x^2) = 2$

Now since the limit exists and is the same on both sides, it follows that

$\lim_{x \rightarrow 1} f(x) = 2$

#### Practice

Find the following limits:

1. $\lim_{x\to-3^-}$

2. $\lim_{x\to2^+}$

3. $\lim_{x\to-1^+}$ and $\lim_{x\to-1^-}$

4. $\lim_{x\to-1}$

5. $\lim_{x\to-2^-}$ and $\lim_{x\to5^+}$

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Find the following limits based on the equation:

Hint: Graph the equations or look at a table.

1. $\lim_{x\to2^+}\frac{-x^2 - 2x + 8}{x - 2}=$
2. $\lim_{x\to0^+}\frac{-x^2 + 4x}{x}=$
3. $\lim_{x\to1^+}\frac{4x^2 - x - 3}{x - 1}=$
4. $\lim_{x\to0^+}\frac{x^2 - 4x}{x}=$
5. $\lim_{x\to2^-}\frac{4x^2 - 7x - 2}{x - 2}=$
6. $\lim_{x \to -5^-}\frac{-3x^2 - 13x + 10}{x + 5}=$

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Determine if the limits exist:

1. $g(x)= \begin{cases} 3 ; x \geq -1\\ x + 4 ; x < -1\\ \end{cases}$
2. $h(x)= \begin{cases} -2; x \geq -1\\ -5x + 2 ; x < -1\\ \end{cases}$
3. $g(x)= \begin{cases} -2 ; x = - 2\\ -3x + 3 ; x \not= -2\\ \end{cases}$
4. $g(x)= \begin{cases} -3x - 4 ; x = 3\\ -2x - 1 ; x \not= 3\\ \end{cases}$
5. $f(x)= \begin{cases} -3 ; x = -1\\ -2 ; x \not= -1\\ \end{cases}$