# Find each of the following limits if they exist

(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

• 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. – Martin Sleziak Feb 28 '17 at 6:58

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.

• 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) – Mark Viola Feb 27 '17 at 20:53
• @Dr.MV Good points! I added a little bit - let me know if you have any more thoughts to improve it. – Brevan Ellefsen Feb 27 '17 at 21:01
• (+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 $\infty$ – Mark Viola Feb 27 '17 at 21:11
• we haven't proved the product rule also it doesn't say use the epsilon delta definition it just says find the limits – August Haze Feb 27 '17 at 21:52
• @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 $\lim_{x \to a} fg = +\infty$. – Brevan Ellefsen Feb 27 '17 at 21:55

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)

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

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

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

• why 2? (x+1)/(x-1) – August Haze Feb 27 '17 at 21:49
• $0< x-1 \implies x+1> 2$ – Doug M Feb 27 '17 at 21:57
• oh nice thanks i like your method alot – August Haze Feb 27 '17 at 22:03
• wait the first one goes to 1 not to infinity – August Haze Feb 27 '17 at 22:04
• 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. – Doug M Feb 27 '17 at 22:43