I’m confused on this question

Answer:

[tex]\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\\[/tex]

Step-by-step explanation:

[tex]\lim_{x \to 5}f(x)[/tex]

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:

[tex]f(x)=5-2x[/tex]

[tex]f(5)=5-2(5)=5-10=-5[/tex]

[tex]\lim_{x \to 5}f(x)=-5[/tex]

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

[tex]\lim_{x \to 1^-} f(x)[/tex]

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:

[tex]f(x)=4-x[/tex]

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

[tex]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[/tex]

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:

[tex]\lim_{x \to 1^-} f(x)=3[/tex]

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

[tex]\lim_{x \to 1^+} f(x)[/tex]

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

[tex]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[/tex]

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.

[tex]\lim_{x \to 1^+} f(x)=3[/tex]

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

[tex]\lim_{x \to 1} f(x)[/tex]

Here, the function for x=1 is just:

[tex]f(x)=4[/tex]

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

[tex]\lim_{x \to 1} f(x)=4[/tex]

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