[College Calc 01-02. Limits of Sequences] #02

Let us prove the general formula:

$\boxed{\lim_{n\to\infty}\left(\sqrt{an^2+bn+c}-\sqrt{an^2+b'n+c'}\right) = \frac{b-b'}{2\sqrt{a}}}$

We multiply the numerator and denominator by $\sqrt{an^2+bn+c}+\sqrt{an^2+b'n+c'}$ such that we have

$\lim_{n\to\infty}\frac{(\sqrt{an^2+bn+c}-\sqrt{an^2+b'n+c'})(\sqrt{an^2+bn+c}+\sqrt{an^2+b'n+c'})}{\sqrt{an^2+bn+c}+\sqrt{an^2+b'n+c'}} = \lim_{n\to\infty}\frac{(b-b')n+c-c'}{\sqrt{an^2+bn+c}+\sqrt{an^2+b'n+c'}}.$

Notice, that the denominator can be replaced by $2\sqrt{a}n$ and the numerator by $(b-b')n.$ Then,

$\cdots = \lim_{n\to\infty}\frac{(b-b')n}{2\sqrt{a}n} = \frac{b-b'}{2\sqrt{a}}. ~\square$

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Substituting the values $a=1,\,b=\frac{7}{5},\,c=-2,\,b'=0,\,c'=0,$ we have

$\cdots = \frac{\frac{7}{5}-0}{2\sqrt{1}} = \frac{7}{10}.$

Hence $p+q=7+10=\boxed{17}.$

The function $\sqrt{n^{2}+\frac{7}{5}n-2} - n$ can be rewritten as $\sqrt{\left(n+\frac{7}{10}\right)^{2}-2.49}-n$ using the completing the square method. As $n$ approaches infinity, the $\left(n+\frac{7}{10}\right)^{2}$ term in the square root predominates meaning $\lim_{n \to \infty}\sqrt{\left(n+\frac{7}{10}\right)^{2}-2.49} = n + \frac{7}{10}$ . The final limit is thus $\lim_{n \to \infty}(n+\frac{7}{10} - n) = \frac{7}{10}$ and $p + q = \boxed{17}$

Rewrite the limit as: $\lim_{n \to \infty} \left( n \sqrt{1 + \frac{7}{5n} - \frac{2}{n^2}}- n \right) = \lim_{n \to \infty} \left( n \left( \sqrt{1 + \frac{7}{5n}} - 1 \right) \right)$

as the $2/n^2$ term is negligible.

Using the negative binomial expansion , $(1+x)^{1/2} \approx 1 + \frac{1}{2}x - \frac{1}{2} \frac{-1}{2} \cdot \frac{1}{2!} x^2$ . Substituting $x = \frac{7}{5n}$ :

$= \lim_{n \to \infty} \left( n \left( 1 + \frac{1}{2}\frac{7}{5n} + \frac{1}{8} \frac{7^2}{(5n)^2} - 1 \right) \right)$ $= \lim_{n \to \infty} \left( n \left( \frac{1}{2}\frac{7}{5n} \right) \right) = \frac{7}{10}$

so $p+q = \boxed{17}$ .

$Y$ = $\sqrt{n^2+\frac{7}{5}n -2}-n$

$Y$ = $n(\sqrt{1+\frac{7}{5n}-\frac{2}{n^2}}-1)$

Here $n$ reaches to $∞$

$Y$ = $n((1+\frac {7}{10n} -\frac{1}{n^2})-1)$

Using property when x reaches to $\sqrt{1+x}$ = $1+\frac{x}{2}$

$Y$ = $n(\frac{7}{10n} -\frac{1}{n^2})$

$Y$ = $\frac{7}{10}-\frac{1}{n}$

Here $\frac{1}{n}$ reaches to zero

So $\boxed{Y =\frac{7}{10}}$

Finally $p+q =\boxed{17}$

$\begin{aligned} L & = \lim_{n \to \infty} \left(\sqrt{n^2+\frac 75n-2}-n\right) \\ & = \lim_{n \to \infty} \frac {\left(\sqrt{n^2+\frac 75n-2}-n\right)\blue{\left(\sqrt{n^2+\frac 75n-2}+n\right)}}\blue {\sqrt{n^2+\frac 75n-2}+n} \\ & = \lim_{n \to \infty} \frac {\frac 75n-2}{\sqrt{n^2+\frac 75n-2}+n} & \small \blue{\text{Divide up and down by }n} \\ & = \lim_{n \to \infty} \frac {\frac 75-\frac 2n}{\sqrt{1+\frac 7{5n}-\frac 2{n^2}}+1} \\ & = \frac 7{10} \end{aligned}$

Therefore $p+q = 7+10 = \boxed{17}$ .

$\begin{aligned} \lim_{n\to \infty} \left(\sqrt{n^2+\frac75 n-2}-n \right) &=\lim_{n\to \infty} n\left(\sqrt{1+\frac{7}{5n}-\frac{2}{n^2}}-1 \right) \\ &=\lim_{n\to \infty} n\left(1+\frac12 \left[\frac{7}{5n}-\frac{2}{n^2}\right]-1 \right) \;\;\;{\color{#BBBBBB} \text{using }\sqrt{1+x} = 1+\frac12 x + \cdots}\\ &=\frac{7}{10} \end{aligned}$

giving the answer $\boxed{17}$ .