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$$\begin{equation*} \lim_{n \rightarrow \infty} \frac{\sqrt{n+1} + 3}{\sqrt{n+2} - 4} \end{equation*}$$

I know that I should multiply by the conjugate if I have square roots in either the numerator or the denominator but what if I have square roots in both numerator and denominator?

Could anyone help me please?

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    $\begingroup$ Both the numerator and the denominator are $\sqrt{n}+O(1)$, so... $\endgroup$ – Jack D'Aurizio Sep 1 '18 at 23:33
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HINT

By intuition we have that the $\sqrt n$ terms are dominant on the others and therefore in the limit

$$\frac{\sqrt{(n+1)}+ 3}{\sqrt{(n+2)} - 4}\sim \frac{\sqrt n}{\sqrt n}=1$$

To check and formalize rigoursly this fact let consider

$$\frac{\sqrt{(n+1)}+ 3}{\sqrt{(n+2)} - 4}=\frac{\sqrt n}{\sqrt n}\frac{\sqrt{(1+1/n)}+ 3/\sqrt n}{\sqrt{(1+2/n)} - 4/\sqrt n}$$

then take the limit.

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Using equivalents, it's very simple:

$\sqrt{n+1}+3,\:\sqrt{n+2}-4\sim_\infty\sqrt n$, so $$\frac{\sqrt{n+1}+3}{\sqrt{n+2}-4}\sim_\infty\frac{\sqrt n}{\sqrt n}=1.$$

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Just if you want to know more than the limit itself. $$\sqrt{n+a}=\sqrt n \sqrt{1+\frac an}$$ Now, use the binomial expansion or Taylor series to get $$\sqrt{1+\frac an}=1+\frac{a}{2 n}-\frac{a^2}{8 n^2}+O\left(\frac{1}{n^3}\right)$$ $$\frac{\sqrt{n+1} + 3}{\sqrt{n+2} - 4}=\frac{\sqrt n\left(1+\frac{1}{2 n}-\frac{1}{8 n^2}+O\left(\frac{1}{n^3}\right) \right)+3 } {\sqrt n\left(1+\frac{1}{n}-\frac{1}{2 n^2} +O\left(\frac{1}{n^3}\right)\right)-4 }=\frac {\sqrt n +3 +\frac1 {2 \sqrt n}-\frac1 {8 n^{3/2}}+O\left(\frac{1}{n^{5/2}}\right) } {\sqrt n -4 +\frac1 { \sqrt n}-\frac1 {2 n^{3/2}}+O\left(\frac{1}{n^{5/2}}\right) }$$ Now, long division to get $$\frac{\sqrt{n+1} + 3}{\sqrt{n+2} - 4}=1+\frac 7 {\sqrt n}+\frac {55}{2n}+O\left(\frac{1}{n^{3/2}}\right)$$

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  • $\begingroup$ That's a nice limit radiography! $\endgroup$ – gimusi Sep 2 '18 at 7:20
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General rules:

(1). If $\lim_{n\to \infty}A_n=A$ and $\lim_{n\to \infty}B_n=B$ then

(1a). $\lim_{n\to \infty}(A_n+B_n)=A+B.$

(1b). If $B\ne 0$ then $\lim_{n\to \infty} (A_n/B_n)=A/B.$

(2). If $\lim_{n\to \infty}A_n=A\geq 0$ then $\lim_{n\to \infty }\sqrt{A_n}=\sqrt A.$

We have $\lim_{n\to \infty}\frac {\sqrt {n+1}+3}{\sqrt {n+2}-4}=$ $\lim_{n\to \infty}\frac {\sqrt {1+1/n}\;+(3/\sqrt n)}{\sqrt {1+2/n} \;-(4/\sqrt n)}.$

We have $1=\lim_{n\to \infty}(1+1/n)=\lim_{n\to \infty}(1+2/n).$ We have $0=\lim_{n\to \infty}(3/\sqrt n)=\lim_{n\to \infty}(-4/\sqrt n).$

Applying the rules above , we obtain a limit of $1$ for both the numerator and denominator of the given expression, and therefore a limit of $1$ for the whole thing.

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Based on the estimate: $$\color{blue}{\frac{\sqrt{n+1}+3}{\sqrt{n+2}}}<\frac{\sqrt{n+1}+3}{\sqrt{n+2}-4}<\color{red}{\frac{\sqrt{n+2}+3}{\sqrt{n+2}-4}}, n>14 \iff \\ \lim_{n\to\infty}\color{blue}{\frac{\sqrt{n+1}+3}{\sqrt{n+2}}}=\lim_{n\to\infty}\sqrt{1-\frac{1}{n+2}}+\lim_{n\to\infty}\frac{3}{\sqrt{n+2}}=1+0=1;\\ \lim_{n\to\infty}\color{red}{\frac{\sqrt{n+2}+3}{\sqrt{n+2}-4}}=\lim_{n\to\infty}\left(1+\frac{7}{\sqrt{n+2}-4}\right)=1.$$ By the Squeeze Theorem the limit is $1$.

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