Question

Show that lim x->a f(x) = L. Factor |f(x) - L| into |"something"||x - a|, then bound "something" by restricting delta. f(x) = 1/(x+1), a= -2, L= -1. Im having alot of trouble with this trying to figure out what to do after restricting the something. I came up with f(x) = |x+2||1/(x+1)| and if delta is less than 1 I got -3 12 years ago

Answer 1

\[\lim_{x\to-2}\frac{1}{x+1}=-1\]
Just to be clear, you're proving this limit, correct?

Answer 2

yes

Answer 3

Okay, so for your scratch work you start with the epsilon side:
\[\begin{align*}\left|\frac{1}{x+1}-(-1)\right|&<\epsilon\\
\left|\frac{x+2}{x+1}\right|&<\epsilon\\
|x+2|\left|\frac{1}{x+1}\right|&<\epsilon
\end{align*}\]
Setting \(\delta<1\), you guarantee that \(|x-(-1)|=|x+1|<1\), or \(-1<x+1<1\), which tells you that \(0<x+2<2\), or \(|x+2|<2\).

So under this assumption,
\[|x+2|\left|\frac{1}{x+1}\right|<2\left|\frac{1}{x+1}\right|<\epsilon~~\Rightarrow~~|x+1|>\frac{2}{\epsilon}\]
So, you'll have to choose \(\delta=\min\left(1,\dfrac{2}{\epsilon}\right)\).

Answer 4

I think, I've always had a hard time with these proofs...

Answer 5

where did you get \[\left| x - (-1) \right|\], and why is that equal to delta?

Answer 6

From the definition of the limit, you have to show that \(0<|x-a|<\delta\) implies \(|f(x)-L|<\epsilon\). In this case, \(a=-1\).

So what you want to get out of this is that as long as this quantity is less than delta, the rest of the proof follows. From there, we agree to set a bound on \(\delta\) and, by extension, the \(|x+2|\) factor we want to get rid of.

Sorry if that doesn't make sense, the idea is somewhat clear in my mind and I'm not entirely sure how to articulate it.

Answer 7

And to clarify, it's not "equal" to delta, it's strictly less than

Answer 8

ok, thanks man! It makes a little more sense.

Answer 9

You're welcome! Keep in mind, you still have to state this formally as a proof.