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I was doing my homework and I stumbled over this particular exercise. I would've known how to solve it, if it had been the same thing under square root in both cases. $$ \lim_{n\to \infty} \left(\sqrt{4n^2+3n+2}-\sqrt{4n^2+n-1}\right) $$

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    $\begingroup$ Hint : get rid of the square root by multiplying by $$\dfrac{\sqrt{4n^2+3n+2}+\sqrt{4n^2+n-1}}{\sqrt{4n^2+3n+2}+\sqrt{4n^2+n-1}}$$ $\endgroup$ – Alain Remillard Nov 12 '19 at 17:03
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By binomial approximation

$$\sqrt{4n^2+3n+2}=2n\left(1+\frac3{4n}+\frac1{2n^2}\right)^\frac12 \approx 2n+\frac34+\frac1{2n}$$

$$\sqrt{4n^2+n-1}=2n\left(1+\frac1{4n}-\frac1{4n^2}\right)^\frac12 \approx 2n+\frac14-\frac1{4n}$$

therefore

$$\sqrt{4n^2+3n+2}-\sqrt{4n^2+n-1}\approx 2n+\frac34+\frac1{2n}-2n-\frac14+\frac1{4n}=\frac12+\frac3{4n}\to \frac12$$

or as alternative, we can use the standard trick $$A-B=(A-B)\dfrac{A+B}{A+B}=\frac{A^2-B^2}{A+B}$$

to obtain

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

$$=(\sqrt{4n^2+3n+2}-\sqrt{4n^2+n-1})\dfrac{\sqrt{4n^2+3n+2}+\sqrt{4n^2+n-1}}{\sqrt{4n^2+3n+2}+\sqrt{4n^2+n-1}}=$$

$$=\dfrac{4n^2+3n+2-4n^2-n+1}{\sqrt{4n^2+3n+2}+\sqrt{4n^2+n-1}}=$$

$$=\dfrac{2n+3}{\sqrt{4n^2+3n+2}+\sqrt{4n^2+n-1}}=$$

$$=\dfrac{2+\frac3n}{\sqrt{4+\frac3n+\frac2{n^2}}+\sqrt{4+\frac1n-\frac1{n^2}}} \to \frac2 4 = \frac12$$

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We can multiply a function by its conjugate

(multiplied by) $$ (\sqrt{4n^2+3n+2}+\sqrt{4n^2+n-1}) $$ $$ \lim \limits_{n\to +\infty} (\frac{4n^2+3n+2-(4n^2+n-1)}{\sqrt{4n^2+3n+3}+\sqrt{4n^2+n-1}}) $$ Thus it turns out $$ \lim \limits_{n\to +\infty} (\frac{2n+3}{4n})=\frac{1}{2} $$

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  • $\begingroup$ The idea is good but I think there is some issue with the denominator in the final step. How can it be equal to 2n? $\endgroup$ – user Nov 12 '19 at 17:27
  • $\begingroup$ sorry one moment $\endgroup$ – vic165 Nov 12 '19 at 17:28
  • $\begingroup$ You should be more precise on the fact that we multiply both numerator and denominator by $(\sqrt{4n^2+3n+2}+\sqrt{4n^2+n-1})$. Moreover to finish it should be better use that $$\frac{4n^2+3n+2-(4n^2+n-1)}{\sqrt{4n^2+3n+3}+\sqrt{4n^2+n-1}}\sim \frac{2n(+3)}{4n} \to \frac12$$ $\endgroup$ – user Nov 12 '19 at 17:36
  • $\begingroup$ This is my first time writing in MathJax :( I was in a hurry $\endgroup$ – vic165 Nov 12 '19 at 17:39
  • $\begingroup$ That's fine, you can copy and edit form all the material here and learn fast. $\endgroup$ – user Nov 12 '19 at 17:42
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You can multiply both numerator and denominator by $\sqrt{4n^{2}+3n+2}+\sqrt{4n^{2}+n-1}$. Then you will find $$\sqrt{4n^{2}+3n+2}-\sqrt{4n^{2}+n-1}=\frac{4n^{2}+3n+2-4n^{2}-n+1}{\sqrt{4n^{2}+3n+2}+\sqrt{4n^{2}+n-1}} = \frac{2n(1+3/2n)}{2n\bigg{[}\sqrt{1+\frac{3}{4n}+\frac{1}{4n^{2}}}+\sqrt{1+\frac{1}{4n}-\frac{1}{4n^{2}}}\bigg{]}} =\frac{(1+3/2n)}{\bigg{[}\sqrt{1+\frac{3}{4n}+\frac{1}{4n^{2}}}+\sqrt{1+\frac{1}{4n}-\frac{1}{4n^{2}}}\bigg{]}} $$ Thus, we have $$\lim_{n\to \infty}\sqrt{4n^{2}+3n+2}-\sqrt{4n^{2}+n-1} = \lim_{n\to\infty}\frac{(1+3/2n)}{\bigg{[}\sqrt{1+\frac{3}{4n}+\frac{1}{4n^{2}}}+\sqrt{1+\frac{1}{4n}-\frac{1}{4n^{2}}}\bigg{]}} = \frac{1}{2}$$

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

so evaluate

$$\lim_{n\to\infty} \frac{2n+3}{{\sqrt{4n^2+3n+2}+\sqrt{4n^2+n-1}}}=\lim_{n\to\infty} \frac{2n}{{\sqrt{4n^2+3n+2}+\sqrt{4n^2+n-1}}}+\\\lim_{n\to\infty} \frac{3}{{\sqrt{4n^2+3n+2}+\sqrt{4n^2+n-1}}}$$

by which

$$\lim_{n\to\infty} \frac{2n+3}{{\sqrt{4n^2+3n+2}+\sqrt{4n^2+n-1}}}=\lim_{n\to\infty} \frac{2n}{{\sqrt{4n^2+3n+2}+\sqrt{4n^2+n-1}}}$$

where the second term is zero because the numerator is a constant while the denominator contains $\sqrt{4n^2}$. Therefore

\begin{align}\lim_{n\to\infty} \frac{2n+3}{{\sqrt{4n^2+3n+2}+\sqrt{4n^2+n-1}}}&=\lim_{n\to\infty} \frac{2n}{{2n\sqrt{1+\frac{3}{4n}+\frac{2}{4n^2}}+2n\sqrt{1+\frac{1}{4n}-\frac{1}{4n^2}}}}\\&=\lim_{n\to\infty} \frac{1}{{\sqrt{1+\frac{3}{4n}+\frac{2}{4n^2}}+\sqrt{1+\frac{1}{4n}-\frac{1}{4n^2}}}}\\&=\frac{1}{2} \end{align}

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    $\begingroup$ We can simply say $$\lim_{n\to\infty} \frac{2n}{{2n\sqrt{1+\frac{3}{4n}+\frac{2}{4n^2}}+2n\sqrt{1+\frac{1}{4n}-\frac{1}{4n^2}}}}=\lim_{n\to\infty} \frac{1}{{\sqrt{1+\frac{3}{4n}+\frac{2}{4n^2}}+\sqrt{1+\frac{1}{4n}-\frac{1}{4n^2}}}}=\frac12$$ $\endgroup$ – user Nov 12 '19 at 17:46
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Hint: $$ (1\pm f(x))^n \approx 1 \pm nf(x) $$ for $f(x) \to 0$.

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  • $\begingroup$ Maybe you are referring to $(1\pm f(x))^n \approx 1 \pm nf(x)$? $\endgroup$ – user Nov 12 '19 at 17:20
  • $\begingroup$ Yes. I made the edit. Thanks. $\endgroup$ – Dinno Koluh Nov 12 '19 at 17:25
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    $\begingroup$ That's indeed binomial approximation! $\endgroup$ – user Nov 12 '19 at 17:26
  • $\begingroup$ Yes, a general case. $\endgroup$ – Dinno Koluh Nov 12 '19 at 17:32
  • $\begingroup$ That's a good hint for these kind of limits, in particular for $n\neq \frac12$. $\endgroup$ – user Nov 12 '19 at 17:33

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