Question

Consider the function g(x) = StartLayout Enlarged left-brace first row negative 9, x less-than 11 second row 7, x greater-than-or-equal-to 11 EndLayout.

What is Limit of g (x) as x approaches 11, if it exists?

–9
–1
7
DNE

Answer 1

The limit as g(x) approaches 11 doesn't exist.

$\lim_{x \to 11} g(x) = DNE$

It looks like you are working with the following piecewise defined function:

$g(x)=\left \{ {{-9;x<11} \atop {7;x\geq11}} \right.$

There is a rule in calculus that tells us that for a limit to exist, both lateral limits must be the same as x approaches a given value. This is:

$\lim_{x \to a^{-}} f(x)=\lim_{x \to a^{+}} f(x)$

if and only if this is true, then:

$\lim_{x \to a^{-}} f(x)=\lim_{x \to a^{+}} f(x)=\lim_{x \to a} f(x)$

if that condition isn't met, then the limit doesn't exist.

So let's calculate the lateral limits for this function:

$\lim_{x \to 11^{-}} g(x) = -9$

This means that as x approaches 11 from the left, the function will return a value close to -9

Now, let's calculate the lateral limits for this function:

$\lim_{x \to 11^{+}} g(x) = 7$

This means that as x approaches 11 from the right, the function will return a value close to 7

You can better visualize this in the graph of the function which I attached to this answer.

Notice that both lateral limits are different, from the left you got a -9 and from the right you got a 7. Therefore, according to the rule I wrote above, the limit does not exist.

For more information on this topic, you can go to the following link:

https://brainly.com/question/17334531?referrer=searchResults

Answer 2

Answer:

D) DNE

Step-by-step explanation:

edge 2021