Question

I’m confused on this question

Answer 1

Answer:

$\lim_{x \to 5}f(x)=-5\\\lim_{x \to 1^-} f(x)=3\\\lim_{x \to 1^+} f(x)=3\\\lim_{x \to 1} f(x)=4$

Step-by-step explanation:

$\lim_{x \to 5}f(x)$

This means "the limit of f(x) as x approaches 5"

In this case, f(x) is a basic linear function while x > 1, so a limit is not necessary but it can still be used as it is here. For this one, you can just calculate f(5), and that is:

$f(x)=5-2x$

$f(5)=5-2(5)=5-10=-5$

$\lim_{x \to 5}f(x)=-5$

===================

$\lim_{x \to 1^-} f(x)$

This one is a little different. It means "the limit of f(x) as x approaches 1 from the negative (left) direction"

The negative direction of course means x will be less than 1, so we'll use this function:

$f(x)=4-x$

It's easier if you graph it, but you can also do it like this:

$f(0)=4+0=4\\f(0.5)=4-0.5=3.5\\f(0.75)=4-0.75=3.25\\f(0.9)=4-0.9=3.1\\f(0.99)=4-0.99=3.01$

There's a clear pattern there. You can see that f(1) should be 3, but this function only applies while x is less than 1 so it can never reach 3. It will get infinitely close, and that can be shown with the limit:

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

===================

$\lim_{x \to 1^+} f(x)$

This one works the same way, but from the positive (right) direction.

$f(2)=5-4=1\\f(1.5)=5-3=2\\f(1.25)=5-2.5=2.5\\f(1.1)=5-2.2=2.8\\f(1.01)=5-2.02=2.98\\f(1.001)=5-2.002=2.998$

It will also get infinitely close to 3 when x = 1, but won't ever reach it because this function only applies when x is greater than 1.

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

===================

$\lim_{x \to 1} f(x)$

Here, the function for x=1 is just:

$f(x)=4$

It's given already, so the limit is just:

$\lim_{x \to 1} f(x)=4$