Calculate: $\lim _{x\to \infty }\left(\frac{\sqrt{x^3+4}+\sin \left(x\right)}{\sqrt{x^3+2x^2+7x+11}}\right)$. Calculate:
$$\lim _{x\to \infty }\left(\frac{\sqrt{x^3+4}+\sin \left(x\right)}{\sqrt{x^3+2x^2+7x+11}}\right)$$

Here's my attempt:
I first tried to the statement up. I first let $f(x)=\left(\frac{\sqrt{x^3+4}+\sin \left(x\right)}{\sqrt{x^3+2x^2+7x+11}}\right)$, $f_1(x)=\left(\frac{\sqrt{x^3+4}}{\sqrt{x^3+2x^2+7x+11}}\right)$ and $f_2(x)=\left(\frac{\sin \left(x\right)}{\sqrt{x^3+2x^2+7x+11}}\right)$. We know that $\lim_{x \to \infty}f(x)=\lim_{x \to \infty}f_1(x)+\lim_{x \to \infty}f_2(x)$. Now we can find the limit of $f_1(x)$ and $f_2(x)$, which will give us our solution.Therefore, we have:
\begin{align}
\lim_{x \to \infty}f_1(x)&amp;=\lim_{x \to \infty}\left(\frac{\sqrt{x^3+4}}{\sqrt{x^3+2x^2+7x+11}}\right)\\ 
&amp;= {\sqrt{\lim_{x \to \infty} \frac{x^{3} + 4}{x^{3} + 2 x^{2} + 7 x + 11}}} \\
&amp;= \sqrt{\lim_{x \to \infty} \frac{1 + \frac{4}{x^{3}}}{1 + \frac{2}{x} + \frac{7}{x^{2}} + \frac{11}{x^{3}}}} \\
&amp;= \sqrt{\frac{\lim_{x \to \infty}\left(1 + \frac{4}{x^{3}}\right)}{\lim_{x \to \infty}\left(1 + \frac{2}{x} + \frac{7}{x^{2}} + \frac{11}{x^{3}}\right)}} \\
&amp;= \ ... \\
&amp;= 1
\end{align}
\begin{align}
\lim_{x \to \infty}f_2(x)&amp;=\lim_{x \to \infty}\left(\frac{\sin \left(x\right)}{\sqrt{x^3+2x^2+7x+11}}\right) \\
&amp;= \ ... \\
&amp;= 0
\end{align}
For the limit of $f_2(x)$, I left out the actuall working out but you can either use the squeeze theorem or the fact that since $\sin(x)$ is divergent and the denominator converges to $0$, which means that the overall function will also converge to $0$. According to the Squeeze Theorem, since $- \frac{1}{\sqrt{x^{3} + 2 x^{2} + 7 x + 11}} \leq \frac{\sin{\left(x \right)}}{\sqrt{x^{3} + 2 x^{2} + 7 x + 11}} \leq \frac{1}{\sqrt{x^{3} + 2 x^{2} + 7 x + 11}}$ and both the functions on the left and right converge to $0$, so does $f_2(x)$.
Therefore: $$\lim _{x\to \infty }\left(\frac{\sqrt{x^3+4}+\sin \left(x\right)}{\sqrt{x^3+2x^2+7x+11}}\right)=1$$
However, I'm not sure if this is $100%$ since neither Symbolab nor EMathHelp could determine the solution.
When $x$ is very large the most dominant term in Num. is $x^{3/2}$ and so is in the Den. Hence  the limit as $x$ tends to $\infty$ is 1.
Result from WolframAlpha for Verification