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

Question

Update:

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

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!

Answer 1

$$=\frac{\sqrt{2t^2-1}-1}{t^2-1}\sqrt[3]{4t^3-3t}+\frac{\sqrt[3]{4t^3-3t}-1}{t^2-1}$$ 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.

Answer 2

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.

Answer 3

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.

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

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

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

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.

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

Using your manipulations, we have

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

Answer 4

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} $

Answer 5

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 $$ \frac{1 +10-68+200-256+128}{(1+1+1+1+1+1)}=\frac{15}{6}=\frac52. $$

Answer 6

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:

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

$$(1+x)^n=1+nx+o(x)$$

Thus:

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

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

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

NOTE

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

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

Answer 7

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} $

Answer 8

Limits to infinity are easier, so:

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