**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!