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


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


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!
$$=\frac{\sqrt{2t^2-1}-1}{t^2-1}\sqrt[3]{4t^3-3t}+\frac{\sqrt[3]{4t^3-3t}-1}{t^2-1}$$ 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.
Limits to infinity are easier, so:

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