Need help with a limit to infinity involving a radical with indeterminate form (stuck in the factoring)

Question

this is my first time on Math Exchange, I searched around the site and could not find a question for this math problem so I do not believe that I am asking a previously asked question, if I am please let me know....any way onto the question. (I apologize if the post is long, please let me know if posts should be shorter)

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Ok so I am currently taking Calculus I and we were given a review sheet for our first exam, I was able to figure out pretty much all of the questions on it save this one I am about to show you. I understand how to solve the question, but I keep getting stuck on the factoring think any of you can show me where I am going wrong and what I should do so solve it? (By the way we are not allowed to use l'hospital's rule or this would be too simple)

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$$ \lim_{x \rightarrow \infty} \sqrt{4x^6+7x^3+5}-2x^3 $$ Basically by looking at it I knew that this was going to be in an indeterminate form of $$ \infty-\infty $$ So I began factoring: $$ \lim_{x \rightarrow \infty} = \sqrt{4x^6+7x^3+5}-2x^3 * \qquad \frac{\sqrt{4x^6+7x^3+5}+2x^3}{\sqrt{4x^6+7x^3+5}+2x^3} \\= \qquad \frac{4x^6+7x^3+5-4x^6}{\sqrt{4x^6+7x^3+5}+2x^3}=\qquad \frac{7x^3+5}{\sqrt{4x^6+7x^3+5}+2x^3} $$ Kinda got stuck there....so can anyone be of help to me? It would be much appreciated!

Thanks!

EDIT: Thank you everybody for the quick replies and help, dividing what was left by x^3 put me on the right track to getting all of your answers of 7/4. Thank you much!

Answer 1

If you multiply numerator and denominator by $1/x^3$ you will get $$\lim_{x\to\infty}{7+5/x^3\over\sqrt{4+7/x^3+5/x^6}+2}=\frac 7{\sqrt4+2}=\frac74.$$

Answer 2

Consider writing $$ \sqrt{4x^{6}+7x^{3}+5} = x^{3}\sqrt{4+7x^{-3}+5x^{-6}}. $$

Answer 3

We can use the Taylor series for $\sqrt{1+u}=1+\frac{u}{2}+O(u^2)$. Then we compute: $$ \begin{align*} \sqrt{4x^6+7x^3+5}&=2x^3\sqrt{1+\frac{7x^3+5}{4x^6}}\\ &=2x^3\left(1+\frac{7x^3+5}{8x^6}+O(x^{-6})\right)\\ &=2x^3+\frac{7}{4}+O(x^{-3}). \end{align*} $$ Therefore $\lim_{x\to\infty}\sqrt{4x^6+7x^3+5}-2x^3=\frac{7}{4}$.