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(a) $\lim\limits_{x \to 1^+} \frac{x+1}{x-1}$

(b) $\lim\limits_{x \to 0^+} \lvert{x^3\sin(1/x)}\rvert$

I am in analysis wondering what methods I can use we just proved limits using the definition of limits involving epsilon and delta and it says "find each of the limits if they exist". Maybe someone can help me understand these. I think I can use the squeeze theorem on the second one.

For (a) I just want to be able to put 1 in for x so I multiplied by $\frac{(x-1)}{(x-1)}$ and got $\frac{(x^2-1)}{(x^2-2x+1)}$ then I can put 1 for x, which makes it 0. It doesn't say use the definition.

This is not correct

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    Please, post only one question in one post. Posting several questions in the same post is discouraged and such questions may be put on hold, see [meta](http://meta.math.stackexchange.com/questions/6464/posting-multiple-questions-as-one). – 2017-02-28

2 Answers 2

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Are you required to use $\epsilon-\delta$ proofs? You explanation isn't clear.

I will assume you have proved the product rule already for the first limit. $$\lim_{x \to 1^+} \frac{x+1}{x-1} = \left(\lim_{x \to 1^+} \frac{1}{x-1}\right)\left(\lim_{x \to 1^+} x+1\right) = 2\lim_{x \to 0^+} \frac{1}{x}$$

We can apply the product rule here because we know that $f = \lim_{x \to 1^+} (x+1) = L$ is finite and positive and that $g = \lim_{x \to 1^+} \frac{1}{x-1}$ goes to $+\infty$, and thus $\lim_{x \to 1^+} fg = +\infty$. Accordingly, for a more rigorous proof the above equalities are actually backwards, as we must calculate the individual limits to show that the product rule applies.

For the second you mention the squeeze theorem, so I assume you have proved this already.Simply note that $$0\le\lvert{x^3\sin(1/x)}\rvert \le x^3$$ and take the limit as $x \to 0$ on both sides.

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    For the second problem, $$0\le |x^3\sin(1/x)|\le x^3$$So, the "$-x^3$" term on the right-hand side is superfluous. For the first problem, it is inappropriate to write $\lim fg=\lim f\times \lim g$ in general. However, if $\lim f=L>0$ and $\lim g=\infty$, then $\lim fg=\infty$. You might consider tidying these up a bit after which (+1) – 2017-02-27
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    @Dr.MV Good points! I added a little bit - let me know if you have any more thoughts to improve it. – 2017-02-27
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    (+1) I would have preferred that you didn't write the product rule, but rather just noted that $\lim x+1=2$ while $\lim 1/(x-1)=\infty$ together imply the limit of the product is $\infty$ – 2017-02-27
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    we haven't proved the product rule also it doesn't say use the epsilon delta definition it just says find the limits – 2017-02-27
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    @AugustHaze well, in that case don't call it the product rule. It's pretty obvious that if $\lim_{x \to a} f = L >0$ and $\lim_{x \to a} g = +\infty$ then $\lim_{x \to a} fg = +\infty$. – 2017-02-27
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    yea that is obvious but with our function i figured the limit to be 0 by multiplying by the conjugate, just a simple little algebra trick we learned in college algebra... – 2017-02-27
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    @AugustHaze the limit is most definitely not 0. A simple graph will confirm that the right limit is $+\infty$ and the left limit is $-\infty$ – 2017-02-27
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    @AugustHaze if you write up your work in either a comment or your main post I can look through it though. – 2017-02-27
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    I see that it is not 0 – 2017-02-28
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a)

This one is going to infinity, so

$\forall N >0,\exists \delta>0 : 0<(x-1)<\delta \implies f(x) > N$

$\frac {x+1}{x-1} > N$

$\frac {x+1}{x-1} > \frac 2{x-1} > \frac {2}{\delta} > N$

$\delta = \frac {2}{N}$

b)

$\forall \epsilon >0,\exists \delta>0 : |x|<\delta \implies |f(x)| <\epsilon$

$|\sin \frac 1x| \le 1\\ |x^3\sin \frac 1x| \le |x^3|$

$\delta = \min (1,\epsilon^{\frac 13})$

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    why 2? (x+1)/(x-1) – 2017-02-27
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    $0< x-1 \implies x+1> 2$ – 2017-02-27
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    oh nice thanks i like your method alot – 2017-02-27
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    wait the first one goes to 1 not to infinity – 2017-02-27
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    No, I think I have proven that it goes to infinity. Suppose $x = 1.01, f(x) = \frac {2.01}{0.01} = 201$ and the closer $x$ get to $1,$ the bigger $f(x)$ gets. – 2017-02-27