# How do I solve this:

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. $$\lim_{n\to \infty} \left(\sqrt{4n^2+3n+2}-\sqrt{4n^2+n-1}\right)$$

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

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

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

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

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

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

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

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

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

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 $$\lim \limits_{n\to +\infty} (\frac{2n+3}{4n})=\frac{1}{2}$$

• 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? – user Nov 12 '19 at 17:27
• sorry one moment – vic165 Nov 12 '19 at 17:28
• 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 $$\frac{4n^2+3n+2-(4n^2+n-1)}{\sqrt{4n^2+3n+3}+\sqrt{4n^2+n-1}}\sim \frac{2n(+3)}{4n} \to \frac12$$ – user Nov 12 '19 at 17:36
• This is my first time writing in MathJax :( I was in a hurry – vic165 Nov 12 '19 at 17:39
• That's fine, you can copy and edit form all the material here and learn fast. – user Nov 12 '19 at 17:42

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 $$\lim_{n\to \infty}\sqrt{4n^{2}+3n+2}-\sqrt{4n^{2}+n-1} = \lim_{n\to\infty}\frac{(1+3/2n)}{\bigg{[}\sqrt{1+\frac{3}{4n}+\frac{1}{4n^{2}}}+\sqrt{1+\frac{1}{4n}-\frac{1}{4n^{2}}}\bigg{]}} = \frac{1}{2}$$

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

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}

• We can simply say $$\lim_{n\to\infty} \frac{2n}{{2n\sqrt{1+\frac{3}{4n}+\frac{2}{4n^2}}+2n\sqrt{1+\frac{1}{4n}-\frac{1}{4n^2}}}}=\lim_{n\to\infty} \frac{1}{{\sqrt{1+\frac{3}{4n}+\frac{2}{4n^2}}+\sqrt{1+\frac{1}{4n}-\frac{1}{4n^2}}}}=\frac12$$ – user Nov 12 '19 at 17:46

Hint: $$(1\pm f(x))^n \approx 1 \pm nf(x)$$ for $$f(x) \to 0$$.

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