$$\lim_{t\to1} \frac {\sqrt {2t^2-1}\sqrt[3]{4t^3-3t}-1}{t^2-1}$$

First of all, I am grateful to you for all the answers you have given me.

I want to ask MSE to confirm the correctness of the alternate solution and its mistake.

I worked so hard for solve this limit without L'Hôpital. I tried to solve this limit myself. Because, I like it. And I trust MSE. Because, MSE is always the real teacher for me. Please, teach me., my mistakes.

$$\begin{align}&\lim_{t \to 1}\frac {\sqrt{2t^2-1}×\sqrt[3]{4t^3-3t}-1}{t^2-1}\\\\&=\lim_{t \to 1} \frac {\sqrt[3]{4t^3-3t}-\frac{1}{\sqrt{2t^2-1}}}{\frac{t^2-1}{\sqrt{2t^2-1}}}\\\\&=\lim_{t \to 1}\frac{4t(t^2-1)+t-\frac{1}{(2t^2-1)×\sqrt{2t^2-1}}}{(t^2-1)×\left[ \frac{\sqrt[3]{(4t^3-3t)^2}}{\sqrt{2t^2-1}}+\frac{\sqrt[3]{4t^3-3t}}{2t^2-1}+\frac{1}{(2t^2-1)×\sqrt{2t^2-1}}\right]}\\\\&=\lim_{t \to 1}\frac{4t+\frac{t(2t^2-1)×\sqrt{2t^2-1}-1}{(2t^2-1)×\sqrt{2t^2-1}×(t^2-1)}}{\left[ \frac{\sqrt[3]{(4t^3-3t)^2}}{\sqrt{2t^2-1}}+\frac{\sqrt[3]{4t^3-3t}}{2t^2-1}+\frac{1}{(2t^2-1)×\sqrt{2t^2-1}}\right]}\\\\&=\lim_{t \to 1}\frac{4t}{\left[ \frac{\sqrt[3]{(4t^3-3t)^2}}{\sqrt{2t^2-1}}+\frac{\sqrt[3]{4t^3-3t}}{2t^2-1}+\frac{1}{(2t^2-1)×\sqrt{2t^2-1}}\right] }\\\\&\qquad\qquad+\lim_{t \to 1}\frac{\frac{t(2t^2-1)×\sqrt{2t^2-1}-1}{(2t^2-1)×\sqrt{2t^2-1}×(t^2-1)}}{\left[ \frac{\sqrt[3]{(4t^3-3t)^2}}{\sqrt{2t^2-1}}+\frac{\sqrt[3]{4t^3-3t}}{2t^2-1}+\frac{1}{(2t^2-1)×\sqrt{2t^2-1}}\right] }\\\\&= \frac{4}{3}+\frac{1}{3}\lim_{t \to 1}\frac{t(2t^2-2)×\sqrt{2t^2-1}+t×\sqrt{2t^2-1}-1}{(t^2-1)×(2t^2-1)×\sqrt{2t^2-1}}\\\\&=\frac{4}{3}+\frac 13\lim_{t \to 1}\frac{2t}{2t^2-1}+\frac 13\lim_{t \to 1}\frac{t×\sqrt{2t^2-1}-1}{(t^2-1)(2t^2-1)\sqrt{2t^2-1}}\\\\&=\frac{4}{3}+\frac 23+\frac 13\lim_{t \to 1}\frac{2t^4-t^2-1}{(t^2-1)(2t^2-1)\sqrt{2t^2-1}×(\sqrt{2t^4-t^2}+1)}\\\\&=2+\frac 13 \lim_{t \to 1}\frac{(t^2-1)(2t^2+1)}{(t^2-1)(2t^2-1)\sqrt{2t^2-1}×(\sqrt{2t^4-t^2}+1)}\\\\&=2+\frac 13 \lim_{t \to 1}\frac{(2t^2+1)}{(2t^2-1)\sqrt{2t^2-1}×(\sqrt{2t^4-t^2}+1)}\\\\&=2+\frac 13×\frac{3}{2}=2+\frac 12=\frac 52.\end{align}$$

I doubt that I have correctly applied the limit rules. Did I apply all the limit rules correctly and is the solution correct..?

Thank you!

  • $\begingroup$ Do you have a specific reason to avoid L’Hopital? $\endgroup$ – abiessu Dec 18 '17 at 0:21
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    $\begingroup$ $\lim\limits_{x\to2}x^2=4$ and $\lim\limits_{x\to-2}x^2=4$. Is that a contradiction? $\endgroup$ – zipirovich Dec 18 '17 at 1:01
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    $\begingroup$ Your solution is correct, but involves unnecessarily complicated expressions. In a typical exam most people would deduct marks for this solution because to figure out that it is correct requires some amount of time. You have overworked for a simple problem. $\endgroup$ – Paramanand Singh Dec 21 '17 at 5:20
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    $\begingroup$ If you study rules of derivatives (along with proofs) you will at once recognize the technique used by Rene Schipperus in his answer which is based on product rule for derivatives. The technique helps to break down a complicated expression into multiple less complicated expressions. And you should apply it as and when needed. $\endgroup$ – Paramanand Singh Dec 21 '17 at 5:31
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    $\begingroup$ It's hard to go through all these formulas lacking explanations of what you are doing. For instance, I can't see where the line after $=\frac{4}{3}+\frac{1}{3}$ comes from. $\endgroup$ – egreg Dec 21 '17 at 10:35

You did a hard work, but it should be better to add some justifications when you use $$\lim_{t\to 1}(f(t)+g(t))=\lim_{t\to 1}f(t)+\lim_{t\to 1}g(t)\tag1$$ $$\lim_{t\to 1}f(t)g(t)=\lim_{t\to 1}f(t)\lim_{t\to 1}g(t)\tag2$$ since these do not always hold.

$$\begin{align}\lim_{t \to 1}\frac {\sqrt{2t^2-1}×\sqrt[3]{4t^3-3t}-1}{t^2-1}&=\lim_{t \to 1} \frac {\sqrt[3]{4t^3-3t}-\frac{1}{\sqrt{2t^2-1}}}{\frac{t^2-1}{\sqrt{2t^2-1}}}\\\\&=\lim_{t \to 1}\frac{4t(t^2-1)+t-\frac{1}{(2t^2-1)×\sqrt{2t^2-1}}}{(t^2-1)×\left[ \frac{\sqrt[3]{(4t^3-3t)^2}}{\sqrt{2t^2-1}}+\frac{\sqrt[3]{4t^3-3t}}{2t^2-1}+\frac{1}{(2t^2-1)×\sqrt{2t^2-1}}\right]}\\\\&=\lim_{t \to 1}\frac{4t+\frac{t(2t^2-1)×\sqrt{2t^2-1}-1}{(2t^2-1)×\sqrt{2t^2-1}×(t^2-1)}}{\left[ \frac{\sqrt[3]{(4t^3-3t)^2}}{\sqrt{2t^2-1}}+\frac{\sqrt[3]{4t^3-3t}}{2t^2-1}+\frac{1}{(2t^2-1)×\sqrt{2t^2-1}}\right]}\end{align}$$

This is correct. (It seems that you used $a-b=\frac{a^3-b^3}{a^2+ab+b^2}$)

$$\begin{align}&\lim_{t \to 1}\frac{4t+\frac{t(2t^2-1)×\sqrt{2t^2-1}-1}{(2t^2-1)×\sqrt{2t^2-1}×(t^2-1)}}{\left[ \frac{\sqrt[3]{(4t^3-3t)^2}}{\sqrt{2t^2-1}}+\frac{\sqrt[3]{4t^3-3t}}{2t^2-1}+\frac{1}{(2t^2-1)×\sqrt{2t^2-1}}\right]}\\\\&=\lim_{t \to 1}\frac{4t}{\left[ \frac{\sqrt[3]{(4t^3-3t)^2}}{\sqrt{2t^2-1}}+\frac{\sqrt[3]{4t^3-3t}}{2t^2-1}+\frac{1}{(2t^2-1)×\sqrt{2t^2-1}}\right]}\\\\&\qquad\qquad\quad+\lim_{t \to 1}\frac{\frac{t(2t^2-1)×\sqrt{2t^2-1}-1}{(2t^2-1)×\sqrt{2t^2-1}×(t^2-1)}}{\left[ \frac{\sqrt[3]{(4t^3-3t)^2}}{\sqrt{2t^2-1}}+\frac{\sqrt[3]{4t^3-3t}}{2t^2-1}+\frac{1}{(2t^2-1)×\sqrt{2t^2-1}}\right]}\\\\&=\frac{4}{3}+\frac{1}{3}\lim_{t \to 1}\frac{t(2t^2-2)×\sqrt{2t^2-1}+t×\sqrt{2t^2-1}-1}{(t^2-1)×(2t^2-1)×\sqrt{2t^2-1}}\end{align}$$

This is correct, but it should be better to add some explanations about why you can use $(1)(2)$ here. I think using $(1)(2)$ without any justifications is not good.

You can avoid using $(1)(2)$.

Using your manipulations, we have

$$\begin{align}&\lim_{t \to 1}\frac{4t+\frac{t(2t^2-1)×\sqrt{2t^2-1}-1}{(2t^2-1)×\sqrt{2t^2-1}×(t^2-1)}}{\left[ \frac{\sqrt[3]{(4t^3-3t)^2}}{\sqrt{2t^2-1}}+\frac{\sqrt[3]{4t^3-3t}}{2t^2-1}+\frac{1}{(2t^2-1)×\sqrt{2t^2-1}}\right]}\\\\&=\lim_{t \to 1}\frac{4t+\frac{t(2t^2-2)×\sqrt{2t^2-1}+t×\sqrt{2t^2-1}-1}{(t^2-1)×(2t^2-1)×\sqrt{2t^2-1}}}{\left[ \frac{\sqrt[3]{(4t^3-3t)^2}}{\sqrt{2t^2-1}}+\frac{\sqrt[3]{4t^3-3t}}{2t^2-1}+\frac{1}{(2t^2-1)×\sqrt{2t^2-1}}\right]}\\\\&=\lim_{t \to 1}\frac{4t+\frac{2t}{2t^2-1}+\frac{t×\sqrt{2t^2-1}-1}{(t^2-1)(2t^2-1)\sqrt{2t^2-1}}}{\left[ \frac{\sqrt[3]{(4t^3-3t)^2}}{\sqrt{2t^2-1}}+\frac{\sqrt[3]{4t^3-3t}}{2t^2-1}+\frac{1}{(2t^2-1)×\sqrt{2t^2-1}}\right]}\\\\&=\lim_{t \to 1}\frac{4t+\frac{2t}{2t^2-1}+\frac{2t^4-t^2-1}{(t^2-1)(2t^2-1)\sqrt{2t^2-1}×(\sqrt{2t^4-t^2}+1)}}{\left[ \frac{\sqrt[3]{(4t^3-3t)^2}}{\sqrt{2t^2-1}}+\frac{\sqrt[3]{4t^3-3t}}{2t^2-1}+\frac{1}{(2t^2-1)×\sqrt{2t^2-1}}\right]}\\\\&=\lim_{t \to 1}\frac{4t+\frac{2t}{2t^2-1}+\frac{(t^2-1)(2t^2+1)}{(t^2-1)(2t^2-1)\sqrt{2t^2-1}×(\sqrt{2t^4-t^2}+1)}}{\left[ \frac{\sqrt[3]{(4t^3-3t)^2}}{\sqrt{2t^2-1}}+\frac{\sqrt[3]{4t^3-3t}}{2t^2-1}+\frac{1}{(2t^2-1)×\sqrt{2t^2-1}}\right]}\\\\&=\lim_{t \to 1}\frac{4t+\frac{2t}{2t^2-1}+\frac{(2t^2+1)}{(2t^2-1)\sqrt{2t^2-1}×(\sqrt{2t^4-t^2}+1)}}{\left[ \frac{\sqrt[3]{(4t^3-3t)^2}}{\sqrt{2t^2-1}}+\frac{\sqrt[3]{4t^3-3t}}{2t^2-1}+\frac{1}{(2t^2-1)×\sqrt{2t^2-1}}\right]}\\\\&=\frac{4+2+\frac 32}{1+1+1}=\frac 52\end{align}$$

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    $\begingroup$ By the way hats off to you for typing all that $\mathrm \LaTeX $ and the same compliments to @MathLover also. $\endgroup$ – Paramanand Singh Dec 22 '17 at 9:57
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    $\begingroup$ Also I think the use of limit laws is justified in each step and one need not write them explicitly (who writes derivative laws when differentiating a complicated function? And more importantly who demands such things to be explicitly stated in a solution?). The real issue is dealing with complicated expressions and checking that they are written correctly. The chances of making a clerical mistake (like writing Square root instead of cube root or a plus instead of minus) while writing such stuff is very high. $\endgroup$ – Paramanand Singh Dec 22 '17 at 10:02
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    $\begingroup$ First of all, thank you for your informative and excellent answer, dear mathlove :) And I think, your last steps are more elegant.(+1) $\endgroup$ – MathLover Dec 22 '17 at 10:04

$$=\frac{\sqrt{2t^2-1}-1}{t^2-1}\sqrt[3]{4t^3-3t}+\frac{\sqrt[3]{4t^3-3t}-1}{t^2-1}$$ now multiply by conjugates and you get the right answer.

As for the second part is possible that the other value of $t$ gives the same value as the limit to 1.

  • 1
    $\begingroup$ Where does the $\sqrt{2t^2-2}$ come from? $\endgroup$ – abiessu Dec 18 '17 at 0:59
  • $\begingroup$ @abiessu Typo, sorry. $\endgroup$ – Rene Schipperus Dec 18 '17 at 1:03
  • $\begingroup$ Actually, I dont understood..:( can you add some detail? +1) $\endgroup$ – MathLover Dec 18 '17 at 1:15
  • $\begingroup$ Well, I dont think you understand the method of multiplying by conjugates. $\endgroup$ – Rene Schipperus Dec 18 '17 at 1:23
  • $\begingroup$ Nothing gets simpler than this +1. However instead of using conjugate (and multiplication stuff) we can directly use the (not so) popular formula $(x^{n} - a^{n}) /(x-a) \to na^{n-1}$ as $x\to a$. $\endgroup$ – Paramanand Singh Dec 18 '17 at 4:05

Elegant, I do not know, but a possible solution.

Let $t=x+1$ making the expression to be $$A=\frac{\sqrt{2 x^2+4 x+1} \sqrt[3]{4 x^3+12 x^2+9 x+1}-1}{x (x+2)}$$ Now, using Taylor around $x=0$ or binomial expansion $$\sqrt{2 x^2+4 x+1}=1+2 x-x^2+2 x^3+O\left(x^4\right)$$ $$ \sqrt[3]{4 x^3+12 x^2+9 x+1}=1+3 x-5 x^2+\frac{67 }{3}x^3+O\left(x^4\right)$$ $$\sqrt{2 x^2+4 x+1} \sqrt[3]{4 x^3+12 x^2+9 x+1}=1+5 x+\frac{34 x^3}{3}+O\left(x^4\right)$$ making $$A=\frac{5}{2}-\frac{5 }{4}x+O\left(x^2\right)$$

Concerning the second point, may I suggest you plot thr function for $-1 \leq t \leq 2$ ? You should understand why this number.


The expressions suggest setting $t = \cos x$. The limit transforms to

$\displaystyle \lim_{x \rightarrow 0} \dfrac{1-\sqrt{\cos 2x } \sqrt [3] {\cos 3x }}{\sin^2 x} = \lim_{x \rightarrow 0} \dfrac{1-\sqrt{\cos 2x} +\sqrt{\cos 2x} - \sqrt{\cos 2x } \sqrt [3] {\cos 3x }}{\sin^2 x}$

Now, $\dfrac{1-\sqrt{\cos 2x}}{\sin^2 x} = \dfrac{1-\cos 2x}{\sin^2 x (1+\sqrt{\cos 2x})} = \dfrac{2}{1+\sqrt{\cos 2x}} \rightarrow 1$ as $x \rightarrow 0$

and $\dfrac{\sqrt{\cos 2x} - \sqrt{\cos 2x } \sqrt [3] {\cos 3x }}{\sin^2 x} = \dfrac{\sqrt{\cos 2x} \left(1-\sqrt [3] {\cos 3x } \right)}{\sin^2 x}$

$=\dfrac{\sqrt{\cos 2x} \left(1-\cos 3x \right)}{\sin^2 x \left(1+\sqrt [3] {\cos 3x }+ \left(\sqrt [3] {\cos 3x } \right)^2 \right)}$

$=\dfrac{2 \sqrt{\cos 2x} \sin^2 \dfrac{3x}{2} }{\sin^2 x \left(1+\sqrt [3] {\cos 3x }+ \left(\sqrt [3] {\cos 3x } \right)^2 \right)}$

$=\displaystyle \lim_{x \rightarrow 0} \dfrac{1-\cos 2x \cos 3x }{\sin^2 x (1+\sqrt{\cos 2x \cos 3x })} \rightarrow \dfrac{3}{2}$ as $x \rightarrow 0$

Hence the limit equals $1+\dfrac{3}{2} = \dfrac{5}{2} $

  • 1
    $\begingroup$ This one is not simpler compared to the original limit. The algebraic limit is easily handled by methods of algebra in 3,4 steps. And your trick works only because the expression was chosen in that manner. What would you do if $4t^3-3t$ was changed to $5t^4-4t$? $\endgroup$ – Paramanand Singh Dec 18 '17 at 4:21
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    $\begingroup$ I am giving another perspective, with no claims to it being more elegant or superior or robust. This particular problem seems to have been set using trig substitutions in mind. Although while writing it out fully it may seem long, but with enough practice trig limits are especially amenable to mental calculation. $\endgroup$ – Hari Shankar Dec 18 '17 at 4:30
  • $\begingroup$ I have not given a vote rather just a comment abt the solution and I don't agree that trig limits are especially amenable to mental calculations compared to algebraic limits. Problems such as these are routine whether they be trigonometric or algebraic and it does not matter which approach you choose. $\endgroup$ – Paramanand Singh Dec 18 '17 at 4:38
  • $\begingroup$ In that case your comment is irrelevant and immaterial $\endgroup$ – Hari Shankar Dec 18 '17 at 4:42
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    $\begingroup$ Perhaps you misinterpreted my last comment. I wanted to say that these problems are easily handled by their natural approaches and there is no need to convert an algebraic problem into a trigonometric one and vice versa (noting that such conversions may not always be possible). But I guess you are more efficient with trig identities than algebraic ones, so be it. $\endgroup$ – Paramanand Singh Dec 18 '17 at 5:04

Using the equality $$ x-1=\frac{x^6-1}{1+x+x^2+x^3+x^4+x^5}, $$ we get (using Wolfram Alpha to expand and then to divide) \begin{align} \frac{\sqrt {2t^2-1}\sqrt[3]{4t^3-3t}-1}{t^2-1}&=\frac {({2t^2-1})^3({4t^3-3t})^2-1}{(t^2-1)\sum_{k=0}^5( {2t^2-1})^{k/2}({4t^3-3t})^{k/3}}\\ \ \\ &=\frac{-1 - 9 t^2 + 78 t^4 - 268 t^6 + 456 t^8 - 384 t^{10} + 128 t^{12}}{(t^2-1)\sum_{k=0}^5( {2t^2-1})^{k/2}({4t^3-3t})^{k/3}}\\ \ \\ &=\frac{1 + 10 t^2 - 68 t^4 + 200 t^6 - 256 t^8 + 128 t^{10}}{\sum_{k=0}^5( {2t^2-1})^{k/2}({4t^3-3t})^{k/3}}\\ \ \\ \end{align} For $t=1$ we get $$ \frac{1 +10-68+200-256+128}{(1+1+1+1+1+1)}=\frac{15}{6}=\frac52. $$

  • $\begingroup$ I though of this at first, but it is just too much work. There is a trick to simplify things, see my answer. $\endgroup$ – Rene Schipperus Dec 18 '17 at 0:53
  • $\begingroup$ And $t≈−0.849620116911296?$ $\endgroup$ – MathLover Dec 18 '17 at 0:54
  • $\begingroup$ I dont understood method, but +1. $\endgroup$ – MathLover Dec 18 '17 at 1:05
  • $\begingroup$ @MathLover: all I did is use the formula from the beginning, and the expand and divide polynomials with the help of Wolfram Alpha. Your $-0.84\ldots$ is completely irrelevant. For instance, $\lim_t\to1}t^2=1$. Solving $t^2=1$ and obtaining $-1$ as a solution is absolutely irrelevant to the limit. $\endgroup$ – Martin Argerami Dec 18 '17 at 1:10
  • $\begingroup$ I did the calculation without wolfram and YES they are much simpler. $\endgroup$ – Rene Schipperus Dec 18 '17 at 1:22

Let’s set:

$$\frac {\sqrt {2t^2-1}\sqrt[3]{4t^3-3t}-1}{t^2-1} \quad t=\cos x \quad x\to0$$

$$\frac {\sqrt {2\cos^2x-1}\sqrt[3]{4\cos^3 x-3\cos x}-1}{-\sin^2x} = \frac {\sqrt {\cos 2x}\sqrt[3]{\cos x(2\cos 2x-1)}-1}{-\sin^2x}$$

By Taylor series we have:

$$\cos x=1-\frac{x^2}{2}+o(x^2)$$



$$=\frac {\sqrt {1-2x^2+o(x^2)}\sqrt[3]{(1-\frac{x^2}{2}+o(x^2))(1-4x^2+o(x^2))}-1}{-x^2+o(x^2)}=$$

$$=\frac {\sqrt {1-2x^2+o(x^2)}\sqrt[3]{(1-\frac{9x^2}{2}+o(x^2)}-1}{-x^2+o(x^2)}=$$

$$=\frac {{(1-x^2+o(x^2)}{(1-\frac{3x^2}{2}+o(x^2))}-1}{-x^2+o(x^2)}=\frac {-\frac{5x^2}{2}+o(x^2)}{-x^2+o(x^2)}\to \frac52$$


$$\frac {\sqrt {2t^2-1}\sqrt[3]{4t^3-3t}-1}{t^2-1}=\frac 52 \Rightarrow t≈-0.849620116911296$$

it's a solution of the equation thus the value of $t$ belong to the domain of the function whereas the limit is calculated for a point for which the function is not defined


Let $t = 1+x$. Then

$\begin{array}\\ \dfrac {\sqrt {2t^2-1}\sqrt[3]{4t^3-3t}-1}{t^2-1} &=\dfrac {\sqrt {2x^2+4x+1}\sqrt[3]{4 x^3 + 12 x^2 + 9 x + 1}-1}{x^2+2x}\\ &=\dfrac {(1+2x+O(x^2))(1+3x+O(x^2))-1}{x^2+2x}\\ &=\dfrac {1+5x+O(x^2)-1}{x^2+2x}\\ &=\dfrac {5+O(x)}{x+2}\\ &=\dfrac52+O(x)\\ \end{array} $


Limits to infinity are easier, so:

$$\lim_{t\to1} \frac {\sqrt {2t^2-1}\sqrt[3]{4t^3-3t}-1}{t^2-1} =$$ $$\lim_{u\to0} \frac {\sqrt {2(u+1)^2-1}\sqrt[3]{4(u+1)^3-3(u+1)}-1}{(u+1)^2-1} =$$ $$\lim_{u\to0} \frac {\sqrt {2(u^2+2u+1)-1}\sqrt[3]{4(u^3+3u^2+3u+1)-3(u+1)}-1}{u^2+2u+1-1} =$$ $$\lim_{u\to0} \frac {\sqrt {2u^2+4u+1}\sqrt[3]{4u^3+12u^2+9u+1}-1}{u^2+2u} =$$ $$\lim_{z\to\infty} \frac {\sqrt {2/z^2+4/z+1}\sqrt[3]{4/z^3+12/z^2+9/z+1}-1}{1/z^2+2/z} =$$ $$\lim_{z\to\infty} \frac {\sqrt {2+4z+z^2}\sqrt[3]{4+12z+9z^2+z^3}-z^2}{1+2z} =$$ $$\lim_{z\to\infty} \frac {\sqrt {4+4z+z^2}\sqrt[3]{27+27z+9z^2+z^3}-z^2}{2z} =$$ $$\lim_{z\to\infty} \frac {(z+2)(z+3)-z^2}{2z} =$$ $$\lim_{z\to\infty} \frac {z^2+5z+6-z^2}{2z} =$$ $$\lim_{z\to\infty} \frac {5z+6}{2z} =\frac52.$$

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    $\begingroup$ Your replacements of $2 $ by $4$ and $12z+4$ by $27z+27$ is invalid. Unless you add more details the solution is wrong. - 1 $\endgroup$ – Paramanand Singh Dec 18 '17 at 4:10
  • $\begingroup$ The lower-order terms don’t matter in the limit. For example, $\lim_{z\to\infty} \frac {z+3}{2z} =\lim_{z\to\infty}\frac{z+21}{2z-17}$. Perhaps an explanation of that step might be useful, but the equalities hold and the step is straightforward. $\endgroup$ – Steve Kass Dec 18 '17 at 4:14
  • $\begingroup$ You should explain very clearly by doing the arithmetic / algebra and show that the terms don't matter. The way it is written looks more like hand waving. Thus for example the right approach $(z+3)/2z=(1/2)+3/2z\to 1/2$ and $(z+21)/(2z-17)=(1+21/z)/(2-17/z)\to 1/2$. $\endgroup$ – Paramanand Singh Dec 18 '17 at 4:18
  • $\begingroup$ The expression in question is complicated with roots and a subtraction of $z^2$ at the end is also there. Unless we do the calculations we can't be sure of such things and anyway your procedure is working step by step and each step needs to be self evident / simple. $\endgroup$ – Paramanand Singh Dec 18 '17 at 4:20
  • $\begingroup$ Thanks. I will leave the answer as is, because I expect that my answer together with your comments should make the validity of the step in question clear to readers. I suppose I could note that, for example, $\sqrt {4+4z+z^2}\sqrt[3]{27+27z+9z^2+z^3}-z^2=(z+2+\mathcal o(1))(z+3+\mathcal o(1))$ or something like that, but again, I think the step is straightforward. $\endgroup$ – Steve Kass Dec 18 '17 at 4:24

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