Calculate limit of $\lim_{n \to \infty}4\sqrt{n+3}-\sqrt{n-1}-5\sqrt{n+7}+2\sqrt{n-3}$

Question

Calculate the limit of $\lim_{n \to \infty}4\sqrt{n+3}-\sqrt{n-1}-5\sqrt{n+7}+2\sqrt{n-3}$

I've tried some algebraic tricks but it didn't lead me to the limit.. Hope you'll help, thanks :)

Answer 1

Note that

$$4\sqrt{n+3}-\sqrt{n-1}-5\sqrt{n+7}+2\sqrt{n-3}=$$ $$4(\sqrt{n+3}-\sqrt{n})-(\sqrt{n-1}-\sqrt{n})-5(\sqrt{n+7}-\sqrt{n})+2(\sqrt{n-3}-\sqrt{n})$$

So you just need to find the limit of $$\sqrt{n+a}-\sqrt{n}$$

Answer 2

If the following limit exists $$ \lim_{x\to0}\frac{4\sqrt{1+3x^2}-\sqrt{1-x^2}-5\sqrt{1+7x^2}+2\sqrt{1-3x^2}}{x} $$ then it is equal to your limit (via $n=1/x^2$).

This is just a derivative: $$ 4\frac{6x}{2\sqrt{1+3x^2}}-\frac{-2x}{2\sqrt{1-x^2}} -5\frac{14x}{2\sqrt{1+7x^2}}+2\frac{-6x}{2\sqrt{1-3x^2}} $$ Evaluate at $x=0$.

Answer 3

You probably need Taylor expansion of $\sqrt{1 + x} = 1 + \frac{1}{2}x + o(x)$ as $x \to 0$. Proceed as follows: \begin{align} & 4\sqrt{n + 3} - \sqrt{n - 1} - 5\sqrt{n + 7} + 2\sqrt{n - 3} \\ = & \sqrt{n}\left(4\sqrt{1 + \frac{3}{n}} - \sqrt{1 - \frac{1}{n}} - 5\sqrt{1 + \frac{7}{n}} + 2\sqrt{1 - \frac{3}{n}}\right) \\ = & \sqrt{n}\left[4\left(1 + \frac{3}{2n} + o(1/n)\right) - \left(1 - \frac{1}{2n} + o(1/n)\right) \right. \\ & \left.- 5\left(1 + \frac{7}{2n} + o(1/n)\right) + 2\left(1 - \frac{3}{2n} + o(1/n)\right)\right] \\ = & \sqrt{n}\left(\frac{-28}{2n} + o(1/n) \right) \\ = & -\frac{14}{\sqrt{n}} + o(1/\sqrt{n}) \to 0 \text { as } n \to \infty. \end{align}

Answer 4

Consider any function $$f(x)=\sum_{k=1}^na_k\sqrt{x+b_k}$$ and define $$A=\sum_{k=1}^na_k$$ Then:

• If $A\ne0$: Then, for every $b$, $\sqrt{x+b}\sim\sqrt x$ when $x\to\infty$ hence $f(x)\sim A\sqrt x$, in particular, $$f(x)\to\mathrm{sign}(A)\cdot\infty$$
• If $A=0$: Then, using the identities $$\sqrt{x+b}-\sqrt x=\frac b{\sqrt{b+x}+\sqrt x}$$ for every $b$, and the fact that $A=0$, one gets $$f(x)=\sum_{k=1}^na_k(\sqrt{x+b_k}-\sqrt x)=\sum_{k=1}^n\frac{a_kb_k}{\sqrt{x+b_k}+\sqrt x}=O\left(\frac1{\sqrt x}\right)$$ in particular, $$f(x)\to0$$

Only fact (really) needed: For every $b$, when $x\to\infty$, $\sqrt{x+b}-\sqrt x\to0$.

Answer 5

$$4\sqrt{n+3}-\sqrt{n-1}-5\sqrt{n+7}+2\sqrt{n-3}\\ =(\sqrt{n+3}-\sqrt{n-1})+3(\sqrt{n+3}-\sqrt{n+7})+2(\sqrt{n-3}-\sqrt{n+7})\\ ={3-(-1)\over\sqrt{n+3}+\sqrt{n-1}}+{3(3-7)\over\sqrt{n+3}+\sqrt{n+7}}+{2(-3-7)\over\sqrt{n-3}+\sqrt{n+7}}\\ \to0\qquad\text{as }n\to\infty$$