How do I solve this:

Question

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

Answer 1

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

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

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

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

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

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

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}