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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)&=\lim_{x \to \infty}\left(\frac{\sqrt{x^3+4}}{\sqrt{x^3+2x^2+7x+11}}\right)\\ &= {\sqrt{\lim_{x \to \infty} \frac{x^{3} + 4}{x^{3} + 2 x^{2} + 7 x + 11}}} \\ &= \sqrt{\lim_{x \to \infty} \frac{1 + \frac{4}{x^{3}}}{1 + \frac{2}{x} + \frac{7}{x^{2}} + \frac{11}{x^{3}}}} \\ &= \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)}} \\ &= \ ... \\ &= 1 \end{align}

\begin{align} \lim_{x \to \infty}f_2(x)&=\lim_{x \to \infty}\left(\frac{\sin \left(x\right)}{\sqrt{x^3+2x^2+7x+11}}\right) \\ &= \ ... \\ &= 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.

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  • $\begingroup$ $\lim \frac{f(x)}{g(x)}\neq\frac{\lim f(x)}{\lim g(x)}$ as shown by taking $f(x) = \sin(x)$ and $g(x) = x$ at $+\infty$ $\endgroup$ – charlus Jul 20 '20 at 10:41
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$$\frac{\sqrt{x^3+4}+\sin x}{\sqrt{x^3+2x^2+7x+11}}=\frac{\sqrt{x^3+4}}{\sqrt{x^3+2x^2+7x+11}}+\frac{\sin x}{\sqrt{x^3+2x^2+7x+11}}=$$

$$=\frac{\sqrt{1+\frac4{x^3}}}{\sqrt{1+2\frac1x+\frac7{x^2}+\frac{11}{x^3}}}+\frac{\sin x}{\sqrt{x^3+2x^2+7x+11}}\xrightarrow[x\to\infty]{}1+0=1$$

The second term's limit is zero as it is a function whose limit is zero times a bounded one

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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.

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Result from WolframAlpha for Verification

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