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How to prove? $$\lim_{x\to 2^-}\frac{x^2-16}{x^2-4}=+\infty$$

from this definition $$ \forall {M>0}\;\exists {\delta>0}:\forall{x\in D},\,(0M)$$

I mean if is correct.

I have a problem with $\delta$ because I suppose it should be determined by depending on M but dont' know how to do that

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    XD There is an `automated-theorem-proving` tag?! That is hilarious... – 2017-01-18
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    @SimpleArt: *automated* by *us*, I guess :D – 2017-01-18
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    In a left neighbourhood of $2$ the ratio $\frac{16-x^2}{2+x}$ is close to $3$. Can you prove that $$\lim_{x\to 2^-}\frac{3}{2-x}=\lim_{z\to 0^+}\frac{3}{z}=+\infty$$ through the $\epsilon/\delta$ definition? If so, you can also prove the given claim. – 2017-01-18
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    $\frac{x^2 - 16}{x^2 - 4}= \frac {x^2 -4}{x^2-4} - \frac {12}{x^2 -4}$ On the points where that *is* defined (all but $\pm 2$) that is $1 + \frac {12}{x^2 - 4}$. So those two terms will have the same limit behavior. Ans as we are approaching for $0< x < 2$ we will have $x^2 - 4 < 0$.... – 2017-01-18
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    @JackD'Aurizio Ok, this gets me $\delta=\frac{3}{M}$ and $\frac{3}{x-2}>M$ How make it $f(x)$ again? – 2017-01-18
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    @TheMeff: ask yourself: does this choice of $\delta=\delta(M)$ work also in the original limit / for the original function? If not, how can I fix it? – 2017-01-18

2 Answers 2

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Let $M>4$ (for $0x>2\sqrt{\frac{M-4}{M-1}}$. It now easily follows \begin{align} x&>2\sqrt{\frac{M-4}{M-1}}\\ \sqrt{M-1}x&>2\sqrt{M-4}\\ (M-1)x^2&>4(M-4)\\ M(x^2-4) &>x^2-16\\ \frac{x^2-16}{x^2-4} &> M \end{align}

Which was to be proved.

Doing the last sequence of inequalities in reverse gives how $\delta$ was chosen.

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    thanks a lot, that's what I needed :D – 2017-01-18
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One of my math slogans: Variables should approach zero.

So, letting $x = y+2$,

$\begin{array}\\ \lim_{x\to 2^-}\frac{x^2-16}{x^2-4} &=\lim_{y\to 0^-}\frac{(y+2)^2-16}{(y+2)^2-4}\\ &=\lim_{y\to 0^-}\frac{y^2+4y+4-16}{y^2+4y+4-4}\\ &=\lim_{y\to 0^-}\frac{y^2+4y-12}{y^2+4y}\\ &=\lim_{y\to 0^-}\frac{y^2+4y-12}{y(y+4)}\\ &=\lim_{y\to 0^-}\frac{-12}{4y}\\ &=\lim_{y\to 0^-}\frac{-3}{y}\\ &= +\infty \qquad\text{since } \lim_{y\to 0^-}\frac{1}{y}=-\infty\\ \end{array} $

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    I don't understand where $\lim_{y\to 0^-}\frac{-12}{4y}$came from – 2017-01-19
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    As $y \to 0$, the numerator and $y+4$ go to non-zero values. Those are the values. – 2017-01-19