<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Dismiss
Skip Navigation
You are viewing an older version of this Study Guide. Go to the latest version.

Definition of a Limit

Describes end behavior or behavior of a function at a point.

Atoms Practice
Estimated10 minsto complete
%
Progress
Practice Definition of a Limit
 
 
 
MEMORY METER
This indicates how strong in your memory this concept is
Practice
Progress
Estimated10 minsto complete
%
Practice Now
Turn In
Introduction to Limits

Definition of a Limit

Definition:

The notation   \begin{align*}\lim_{x \rightarrow x_0} f(x) = L\end{align*}    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

.

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
.

Write using limit notation:

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

.

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.

 

.

Find the following limits:

  1. \begin{align*}\lim_{x \rightarrow 0} \frac{5x} {2}\end{align*}
  2. \begin{align*}\lim_{x \rightarrow 4} \sqrt{x}\end{align*}
  3. \begin{align*}\lim_{x \rightarrow 0} \frac{sin x} {x}\end{align*} 
  4. \begin{align*}\lim_{x \rightarrow 0} \frac{3x} {\sqrt{x + 1} - 1}\end{align*} 
  5. \begin{align*}\lim_{x \rightarrow 0} \frac{1-cos x} {x^2}\end{align*}
Click here for the answers.

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?

.

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, \begin{align*}\lim_{x\to0^-} \ne \lim_{x\to0^+}\end{align*}. Therefore  \begin{align*}\lim_{x\to0}\end{align*} 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),

\begin{align*}\lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1^{-}} (3 - x) = (3 - 1) = 2\end{align*}

and for > 1 (limit from the right),

\begin{align*}\lim_{x \rightarrow 1^+} f(x) = \lim_{x \rightarrow 1^+} (3x - x^2) = 2\end{align*}

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

\begin{align*}\lim_{x \rightarrow 1} f(x) = 2\end{align*}


Practice

Find the following limits:

  1. \begin{align*}\lim_{x\to-3^-}\end{align*}

  2. \begin{align*}\lim_{x\to2^+}\end{align*}

  3. \begin{align*}\lim_{x\to-1^+}\end{align*} and \begin{align*}\lim_{x\to-1^-}\end{align*}

  4. \begin{align*}\lim_{x\to-1}\end{align*}

  5. \begin{align*}\lim_{x\to-2^-}\end{align*} and \begin{align*}\lim_{x\to5^+}\end{align*}

.

Find the following limits based on the equation:

Hint: Graph the equations or look at a table.

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

Click here for more help.

.

Determine if the limits exist:

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

Click here for guidance.

Explore More

Sign in to explore more, including practice questions and solutions for Definition of a Limit.
Please wait...
Please wait...