Solving a limit using only special limits and algebric manipulations

Question

I'm wondering how to solve this limit:

$$\lim_{x \to 0^+} \frac{\tan^3((1+x^{\frac 23})^\frac13-1)+\ln(1+\sin^2(x))}{\arctan^2(3x)+5^{x^4}-1}(\sqrt{\frac{1+x+x^2}{x^2}}-\frac 1x)$$

With my actual notions that are:

-Special limits

-A limit of a sum/product/quotient of functions is the sum/product/quotient of limits of those functions if the functions converge(and also if the denominator function doesn't converge to 0 in the case of quotient)

-Basic notions like $+\infty\cdot a=+\infty, a>0$ etc

-Comparison theorem

-Algebric manipulations

Often my teacher does this "trick":

"If we have to calculate: $\lim_\limits{x \to x_0} s(x)c(x)$. Where $s$ is a simple function that we know to be convergent to a non-zero value and $c$ is a complicated functions whom limit is unknown. We can write this: $$ \lim_\limits{x \to x_0} s(x)c(x)=\lim_\limits{x \to x_0} s(x)\lim_\limits{x \to x_0} c(x)$$ If we discover then that: $$\lim_\limits{x \to x_0} c(x)\in \mathbb{R}$$ Then our previous passage is justified. If we discover that: $$\lim_\limits{x \to x_0} c(x)\in \pm \infty$$ Then our previous passage is not justified formally, but it doesn't affect the limit(it's a kind of notation abuse). If we discover that: $$\not\exists \lim_\limits{x \to x_0} c(x)$$ Then our passage is not justified and it may have affected the limit result"

I kinda understood why this works(it's a kind of retrospective justificatin) but i was wondering if there a was a more formal way to describe this, because when i try to do limits I always try to justify all the steps I do and to be formal. However let's go back to the initial limit and to my attempt:

$$\lim_{x \to 0^+} \frac{\tan^3((1+x^{\frac 23})^\frac13-1)+\ln(1+\sin^2(x))}{\arctan^2(3x)+5^{x^4}-1}(\sqrt{\frac{1+x+x^2}{x^2}}-\frac 1x)$$

Let's try to calculate first:

$$\lim_{x \to 0^+} \sqrt{\frac{1+x+x^2}{x^2}}-\frac 1x=\lim_{x \to 0^+} \frac{\sqrt{1+x+x^2}-1}{x}=\lim_{x \to 0^+} \frac{\sqrt{1+x+x^2}-1}{x+x^2}(x+1)$$ Now i use a known special limit: $$\lim_{x \to 0^+} \frac{x+1}{2}=\frac 12$$ Now let's use the trick of my teacher and let's hope that the remaining limit exists otherwise we are at the starting point(this is also why sometimes i'm a bit unsure doing this it feels like a bet): $$\frac 12\lim_{x \to 0^+} \frac{\tan^3((1+x^{\frac 23})^\frac13-1)+\ln(1+\sin^2(x))}{\arctan^2(3x)+5^{x^4}-1}$$

And now i'm stuck because I see many useful special limits that i could apply but it always come to a $$0 \cdot \infty$$ form where i can't apply the "trick". Sometimes I feel i'm overcomplicating everything by being too formal but I really want to understand why I can apply something and I don't want to make it become an automatism before i totally understood it.

Michael Rozenberg · Answer 1 · 2019-12-27T22:12:54.463

2

$$\lim_{x\rightarrow0^+}\left(\sqrt{\tfrac{1+x+x^2}{x^2}}-\tfrac{1}{x}\right)=\lim_{x\rightarrow0^+}\tfrac{\sqrt{1+x+x^2}-1}{x}=\lim_{x\rightarrow0^+}\tfrac{x+x^2}{x(\sqrt{1+x+x^2}+1)}=\lim_{x\rightarrow0^+}\tfrac{1+x}{\sqrt{1+x+x^2}+1}=\frac{1}{2},$$ $$\tan^3\left(\sqrt[3]{1+x^{\frac{2}{3}}}-1\right)=\tan^3\frac{x^{\frac{2}{3}}}{\sqrt[3]{\left(1+x^{\frac{2}{3}}\right)^2}+\sqrt[3]{1+x^{\frac{2}{3}}}+1}=$$ $$=\left(\frac{\tan\frac{x^{\frac{2}{3}}}{\sqrt[3]{\left(1+x^{\frac{2}{3}}\right)^2}+\sqrt[3]{1+x^{\frac{2}{3}}}+1}}{\frac{x^{\frac{2}{3}}}{\sqrt[3]{\left(1+x^{\frac{2}{3}}\right)^2}+\sqrt[3]{1+x^{\frac{2}{3}}}+1}}\right)^3\cdot\frac{\left(\frac{x^{\frac{2}{3}}}{\sqrt[3]{\left(1+x^{\frac{2}{3}}\right)^2}+\sqrt[3]{1+x^{\frac{2}{3}}}+1}\right)^3}{x^2}\cdot x^2\sim 1\cdot\frac{1}{27}\cdot x^2.$$ $$\ln(1+\sin^2x)=\frac{\ln(1+\sin^2x)}{\sin^2x}\cdot\left(\frac{\sin{x}}{x}\right)^2\cdot x^2\sim1\cdot1^2\cdot x^2,$$ $$\arctan^23x=\left(\frac{\arctan3x}{3x}\right)^2\cdot9x^2\sim1^2\cdot9x^2$$ and $$5^{x^4}-1=\frac{e^{x^4\ln5}-1}{x^4\ln5}\cdot x^4\ln5\sim1\cdot x^4\ln5.$$ Now, $$\lim_{x \to 0^+} \frac{\tan^3((1+x^{\frac 23})^\frac13-1)+\ln(1+\sin^2(x))}{\arctan^2(3x)+5^{x^4}-1}\left(\sqrt{\frac{1+x+x^2}{x^2}}-\frac 1x\right)=$$ $$=\lim_{x \to 0^+} \frac{\frac{\tan^3((1+x^{\frac 23})^\frac13-1)}{x^2}+\frac{\ln(1+\sin^2(x))}{x^2}}{\frac{\arctan^23x}{x^2}+\frac{5^{x^4}-1}{x^2}}\lim_{x\rightarrow0^+}\left(\sqrt{\frac{1+x+x^2}{x^2}}-\frac 1x\right)=$$ $$=\frac{\lim\limits_{x\rightarrow0^+}\frac{\tan^3((1+x^{\frac 23})^\frac13-1)}{x^2}+\lim\limits_{x\rightarrow0^+}\frac{\ln(1+\sin^2x)}{x^2}}{\lim\limits_{x\rightarrow0^+}\frac{\arctan^23x}{x^2}+\lim\limits_{x\rightarrow0^+}\frac{5^{x^4}-1}{x^2}}\lim_{x\rightarrow0^+}\left(\sqrt{\frac{1+x+x^2}{x^2}}-\frac 1x\right)=$$ $$=\frac{\frac{1}{27}+1}{9+0}\cdot\frac{1}{2}=\frac{14}{243}.$$ We used the following standard limits: $$\lim_{x\rightarrow0}\frac{\sin{x}}{x}=\lim_{x\rightarrow0}\frac{e^x-1}{x}=\lim_{x\rightarrow0}\frac{\ln(1+x)}{x}=1.$$ Also, we used that $$\tan{x}=\frac{\sin{x}}{\cos{x}}.$$

Also, we used the following useful fact.

Let $f$ is a continuous function and there is $\lim\limits_{x\rightarrow a}g(x).$ Thus: $$\lim_{x\rightarrow a} f(g(x))=f\left(\lim_{x\rightarrow a}g(x)\right).$$

For example, $$\lim_{x\rightarrow0}\sqrt[3]{1+x}=\sqrt[3]{\lim_{x\rightarrow0}(1+x)}=1.$$

edited Dec 27 '19 at 22:12

answered Dec 27 '19 at 19:29

Michael Rozenberg

194,933

I can't use asymptotic equivalences please read well tue question – Kandinskij Dec 27 '19 at 19:30
@Eureka I really calculated the limit. I am ready to explain. Ask your questions. – Michael Rozenberg Dec 27 '19 at 19:32
I can't use asymptotic equivalence ~ because we haven't done that. The list of things I can use is in the question. – Kandinskij Dec 27 '19 at 19:33
@Eureka Is the first line clear now? – Michael Rozenberg Dec 27 '19 at 19:34
yes but why my result is 1/2 and yours 1 where is my mistake? – Kandinskij Dec 27 '19 at 19:40
@Eureka It was my mistake. I fixed. – Michael Rozenberg Dec 27 '19 at 19:41
Ok I got that too what's the next step? – Kandinskij Dec 27 '19 at 19:42
@Eureka I added something again. See now. – Michael Rozenberg Dec 27 '19 at 19:54
I know what you are doing and I also know how to do this too but my problem is not finding and apply special limits. My problem is:" What theorem does allow me to take out some pieces of the function modify them and then taking them back into the function?". I mean you should be able to distribute the limit operator only if you know that all of the functions are convergent but don't know this. – Kandinskij Dec 27 '19 at 19:57
@Eureka I used the limits algebra only. We got a limit: $\frac{14}{243},$ which says that all theorems about limits, which we used are valid. – Michael Rozenberg Dec 27 '19 at 19:59
1

Thank you really much :) , could you tell me more about the trick that my teacher uses, is there a more formal formulation on this kind of operations? – Kandinskij Dec 27 '19 at 21:53
For a fixed $r$ we have $(1+y)^r=1+ry(1+f(y))$ where $\lim_{y\to 0}f(y)=0.$ E.g. when $y=x^{2/3}$ and $r=1/3.$ – DanielWainfleet Dec 27 '19 at 22:53
Can down-voter explain us, why did you do it? – Michael Rozenberg Dec 28 '19 at 13:06

Paramanand Singh · Accepted Answer · 2019-12-28T02:49:39.233

This is a typical example which is designed to intimidate students.

You have already noted that the last factor tends to $1/2$. Without assuming anything about rest of the expression you can move this factor out of the limit to get $$\frac{1}{2}\lim_{x\to 0^{+}} \text{ (rest of the expression)} $$ The next part is to simplify the denominator. Let's write $$\arctan^23x+5^{x^4}-1=(9x^2)\left(\left(\frac{\arctan 3x}{3x}\right)^2+\frac{5^{x^4}-1}{x^4}\cdot\frac{x^2}{9}\right)$$ The expression in large parentheses tends to $$1^2+(\log 5)\cdot 0=1$$ and thus this factor can be safely replaced by $1$ and the denominator simplifies to $9x^2$.

Since the numerator consists of two terms we can now split the expression into two parts the simpler of which is $$\frac{\log(1+\sin^2x)}{9x^2}=\frac{1}{9}\cdot\left(\frac{\sin x} {x} \right) ^2\cdot \frac{\log(1+\sin^2x)}{\sin^2x}$$ and this tends to $(1/9)\cdot 1^2\cdot 1=1/9$. Thus your desired limit equals $$\frac{1}{18}+\frac{1}{18}\lim_{x\to 0^{+}}\frac{\tan^3((1+x^{2/3})^{1/3}-1)}{x^2}$$ The expression under limit above can be written as $$\left(\frac{\tan((1+x^{2/3})^{1/3}-1)}{(1+x^{2/3})^{1/3}-1}\right)^3\cdot\left(\frac{(1+x^{2/3})^{1/3}-1}{(1+x^{2/3})-1}\right)^3$$ which tends to $1^3(1/3)^3=1/27$. Thus the desired limit is $$\frac{1}{18}+\frac{1}{18}\cdot\frac{1}{27}=\frac{14}{243}$$

The trick of your teacher works and has been discussed by me in this post. Apart from this you need the rule for limit of composition of functions.

Theorem: If $$\lim_{x\to a} g(x) =b, \lim_{x\to b} f(x) =L$$ and $g(x) \neq b$ as $x\to a$ then $$\lim_{x\to a} f(g(x)) =L$$

I ask myself why they don't put this on textbook: it would be much more clear :c — Kandinskij, Dec 28 '19 at 09:48
@Eureka: for most textbooks, the goal is to sell large number of copies. Imparting knowledge is secondary motive. — Paramanand Singh, Dec 28 '19 at 09:56

Solving a limit using only special limits and algebric manipulations

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