Please explain this text about limits

[wary solar]

It’s the last bit I don’t understand, starting with “The limit has the form…”.

Why is it dividing the limit of the two functions? Why is it equating them to zero? I don’t get how that follows from what is above it.

-Openstax Calc vol 1 pg 164 if you want to view it on their website

Thanks in advance.

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[lethal swift]

You see how $f(x)=\frac{x^2-1}{x-1}$ has both a denominator and a numerator that is 0 for $x=1$. This 0 in the denominator makes sure that there is discontinuity, however we can still evaluate the limit.

[turbid breach]

Lumberdude #MakeWolfOwner

[wary solar]

Right so, we can’t just plug in 1 to evaluate the limit as x->1, so we need a new strategy. I’m having trouble connecting that to the next part, “The limit has the form…”

[lethal swift]

They just say that you get 0/0 if you just plug in the numbers

That's in essence what it says

Anytime you have that it is in indeterminate form

[wary solar]

OH, I see now, it is just using the numerator as f(x) and the denominator as g(x). I thought it was referring to the f(x) and g(x) at the top of the page

[lethal swift]

Yeah they really dun goofed in doing that

[wary solar]

Thanks for the help

+close