How do you find limits using a graph?

Assuming no points of discontinuity, the limit of a function as x approaches some number, c, can be found by finding the value of the function at x = c.

In other words, for continuous intervals

##lim_(x->c)f(x)=f(c)##

Look at the graph of ##f(x)=x^2## below.

As ##x->2##,the function value approaches 4. Also, f(2) = 4.

So, ##lim_(x->2)f(x)=f(2)=4##

Limits of a function from a graph can get tricky if there are points of discontinuity. Look at the next graph.

The limit of the function is the same as the function value everywhere except at x = -1 and x = 1.

At x = -1 there is a jump discontinuity. The function value is 4 but the limit does not exist ( the limit of f as x approaches -1 from the left is 2 and the limit as x approaches -1 from the right is 4, so the general limit does not exist).

At x = 1 the function has a value of 4 (depicted by the dot at the point (2, 4)). However, the limit of the function is 2 since the function values approach 2 through values greater than 2 as x approaches 1 from the left and the function values approach 2 through values less than 2 as x approaches 1 from the right.

So, in this case

##lim_(x->1)f(x)=2## but ##f(1)=4##

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How do you find limits using a graph?

Assuming no points of discontinuity, the limit of a function as x approaches some number, c, can be found by finding the value of the function at x = c.

In other words, for continuous intervals

##lim_(x->c)f(x)=f(c)##

Look at the graph of ##f(x)=x^2## below.

As ##x->2##,the function value approaches 4. Also, f(2) = 4.

So, ##lim_(x->2)f(x)=f(2)=4##

Limits of a function from a graph can get tricky if there are points of discontinuity. Look at the next graph.

The limit of the function is the same as the function value everywhere except at x = -1 and x = 1.

At x = -1 there is a jump discontinuity. The function value is 4 but the limit does not exist ( the limit of f as x approaches -1 from the left is 2 and the limit as x approaches -1 from the right is 4, so the general limit does not exist).

At x = 1 the function has a value of 4 (depicted by the dot at the point (2, 4)). However, the limit of the function is 2 since the function values approach 2 through values greater than 2 as x approaches 1 from the left and the function values approach 2 through values less than 2 as x approaches 1 from the right.

So, in this case

##lim_(x->1)f(x)=2## but ##f(1)=4##

Leave a Comment

Your email address will not be published. Required fields are marked *

How do you find limits using a graph?

Assuming no points of discontinuity, the limit of a function as x approaches some number, c, can be found by finding the value of the function at x = c.

In other words, for continuous intervals

##lim_(x->c)f(x)=f(c)##

Look at the graph of ##f(x)=x^2## below.

As ##x->2##,the function value approaches 4. Also, f(2) = 4.

So, ##lim_(x->2)f(x)=f(2)=4##

Limits of a function from a graph can get tricky if there are points of discontinuity. Look at the next graph.

The limit of the function is the same as the function value everywhere except at x = -1 and x = 1.

At x = -1 there is a jump discontinuity. The function value is 4 but the limit does not exist ( the limit of f as x approaches -1 from the left is 2 and the limit as x approaches -1 from the right is 4, so the general limit does not exist).

At x = 1 the function has a value of 4 (depicted by the dot at the point (2, 4)). However, the limit of the function is 2 since the function values approach 2 through values greater than 2 as x approaches 1 from the left and the function values approach 2 through values less than 2 as x approaches 1 from the right.

So, in this case

##lim_(x->1)f(x)=2## but ##f(1)=4##


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