Unlike many how-to questions, I would like to know whether my solution is correct and whether all my transformation are equivalent.
Here's the limit I need to compute: $$\underset{x\rightarrow\infty}{\lim}\frac{\sqrt[5]{x^{3}+1}-\sqrt[5]{3x^{4}+2}}{\sqrt[3]{7x^{2}-3x+2}}$$
Step 1
I say $$\sqrt[5]{x^{3}+1}-\sqrt[5]{3x^{4}+2}\sim-\sqrt[5]{3x^{4}+2}, x\rightarrow\infty$$ because $$\underset{x\rightarrow\infty}{\lim}\frac{\sqrt[5]{x^{3}+1}-\sqrt[5]{3x^{4}+2}}{-\sqrt[5]{3x^{4}+2}}=\underset{x\rightarrow\infty}{\lim}\left(-\sqrt[5]{\frac{x^{3}+1}{3x^{4}+2}}+1\right)=0+1=1$$
Thus,
Step 2: $$\underset{x\rightarrow\infty}{\lim}\frac{\sqrt[5]{x^{3}+1}-\sqrt[5]{3x^{4}+2}}{\sqrt[3]{7x^{2}-3x+2}}=\underset{x\rightarrow\infty}{\lim}\frac{-\sqrt[5]{3x^{4}+2}}{\sqrt[3]{7x^{2}-3x+2}}$$
Step 3: $$\underset{x\rightarrow\infty}{\lim}\frac{-\sqrt[5]{3x^{4}+2}}{\sqrt[3]{7x^{2}-3x+2}}=\underset{x\rightarrow\infty}{\lim}\frac{-\sqrt[5]{x^{4}\left(3+2/x^{4}\right)}}{\sqrt[3]{x^{2}\left(7-3/x+2/x^{2}\right)}}=$$
$$=\underset{x\rightarrow\infty}{\lim}-\frac{x^{\frac{4}{5}}\sqrt[5]{\left(3+2/x^{4}\right)}}{x^{\frac{2}{3}}\sqrt[3]{\left(7-3/x+2/x^{2}\right)}}=-\underset{x\rightarrow\infty}{\lim}x^{\frac{2}{15}}*\underset{x\rightarrow\infty}{\lim}\frac{\sqrt[5]{3}}{\sqrt[3]{7}}=-\infty$$
Answer: $-\infty$
Are all the transformation correct and equivalent?
I appreciate any help.