Need help with some tricky $\epsilon, \delta$ proofs

Question

My professor wants $\epsilon\ \delta$ proofs for this... so no using algebraic properities of the limit function or commonly occuring limit identies. I'm hoping that each of these has some algebraic trick to make the proofs of these elegant, although I know my professor makes the problems up himself and his lectures are a bit all over the place so I wouldn't be suprised if the proofs are also.

a) $lim_{x \rightarrow 1} x^3+x^2=2$

b) $lim_{x \rightarrow 4} x^{3/2}=8$

c) $lim_{x \rightarrow 3} \frac{x}{x-2}=3$

I've tried a bunch and haven't made much progress! Thanks everyone !!

Answer 1

$$|x^3+x^2-2|=|(x-1)(x^2+2x+2)|=|x-1||x^2+2x+2|.$$

You can bound $|x^2+2x+2|$ by a constant in a neighborhood of $x=1$ (say $|x^2+2x+2|\le10$ in $[0,2]$), so that

$$|x^2+2x+2|<10|x-1|.$$

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$$|x^{3/2}-8|=|\sqrt x-2||x+2\sqrt x+4|=|x-4|\frac{|x+2\sqrt x+4|}{|\sqrt x+2|}$$ and the fraction can be bounded in a neighborhood of $x=4$.

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$$\left|\frac x{x-2}-3\right|=|x-3|\frac{2}{|x-2|}$$ and the fraction can be bounded in a neighborhood of $x=3$ (of radius $<1$).

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In fact, for any "tamed" function, you will always observe

$$f(x)-f(x_0)\approx|x-x_0|M$$ where $M$ is a constant a little larger than the slope of the function at $x_0$, and

$$\lim_{x\to x_0}f(x)=f(x_0).$$

Answer 2

In each case you begin with.

$\lim_\limits{x\to a} f(x) = L\\ \forall \epsilon>0, \exists \delta > 0 : |x-a|<\delta \implies |f(x) - L| < \epsilon$

e.g.

$\lim_\limits{x\to 1} x^3 + x^2 = 2\\ \forall \epsilon>0, \exists \delta > 0 : |x-1|<\delta \implies |x^3+x^2 - 2| < \epsilon$

Now you want to do some manipulations such that you can factor and $(x-a)$ out of $|f(x) - L|.$

$|x^3+x^2 - 2| = |(x-1)(x^2 + 2x + 2)|$

Now you want to find establish some sort of upper bound for. $|x^2 + 2x + 2|$ (or whatever equivalent factor will be here.)

When $\delta < 1$ then $0<x<2$ and $2<x^2 + 2x + 2<10$

When $\delta = \min (1,\frac {\epsilon}{10})$

$|x-1|<\delta \implies |x^3 + x^2 - 2| < \epsilon$

QED

Answer 3

For the first one, let $x=1+d.$ If $|x-1|<1$ then $|d|<1$ so $$|x^3+x^2-2|=|d^3+4d^2+5d|\leq $$ $$\leq |d|^3+4|d|^2+5|d|\leq$$ $$\leq |d|+4|d|+5|d|=$$ $$=10|d|=10|x-1|.$$

So if $\epsilon>0$ and $\delta=\min (1,\epsilon/10)$ then $$|x-1|<\delta \implies |x^3+x^2-2|\leq 10|x-1|< 10\delta \leq \epsilon.$$