$\forall n\in\mathbb{N}$, $a_n\not=0$, $\lim_{n\to\infty} a_n=0$. Find $\lim_{n\to\infty} \frac{\sqrt{3+a_n}-\sqrt{3}}{a_n}$

Question

I already multiplied by the conjugate so I get $$\lim_{n\to\infty} \frac{1}{\sqrt{3+a_n}+\sqrt{3}}$$ but I dont know how to prove that $\lim_{n\to\infty} \sqrt{3+a_n}=\sqrt{3}$. Also I already tried with the definition with the limit $\frac{1}{2\sqrt{3}}$, but is more hard. Can anyone help me? Also important to say that $a_n$ is a sequence.

Answer 1

Since $\sqrt{\phantom{1}}$ is a continuous function we obtain: $$\lim_{n\rightarrow+\infty}\sqrt{3+a_n}=\sqrt{\lim_{n\rightarrow+\infty}(3+a_n)}=\sqrt3.$$

Answer 2

We know that for positive values, $\sqrt{ \cdot }$ is a continuous function, meaning that the limit of the function is the function of the limit. So $\lim_{n \rightarrow \infty} \sqrt{3+a_n}=\sqrt{\lim_{n \rightarrow \infty}(3+a_n)}$. Find $\lim_{n \rightarrow \infty} 3+a_n$ and you should be good!

Answer 3

$$\begin {align} \lim_{n\to \infty}\sqrt {3+a_n}=\sqrt {\lim_{n\to \infty}\left(3+a_n\right)}=\sqrt {\lim_{n\to \infty }3+\lim_{n\to \infty} a_n}=\sqrt {3+0}=\sqrt 3\end{align} $$

Answer 4

As, $\sqrt{3+x} $ continuous for $x>0$

So, we can take the limit inside of the function.

Using the rule, $$\lim_{n\to\infty} \sqrt{(3+a_n)} = \sqrt{(3+\lim_{n\to\infty} a_n) } = \sqrt{3} $$

Answer 5

Another way.

Let $\epsilon>0$.

Thus, since $\lim\limits_{n\rightarrow+\infty}a_n=0$, there is $N$ such that for any $n>N$ we have $3+a_n\geq0$ and $$|a_{n}|<\sqrt3\epsilon.$$ Thus, $$|\sqrt{3+a_n}-\sqrt3|=\frac{|a_n|}{\sqrt{3+a_n}+\sqrt3}\leq\frac{|a_n|}{\sqrt3}<\epsilon,$$ Which says $$\lim_{n\rightarrow+\infty}\sqrt{3+a_n}=\sqrt3.$$

Indeed, we proved that for any $\epsilon>0$ there is $N$ such that for any $n>N$ we have: $$|\sqrt{3+a_n}-\sqrt3|<e,$$ which by the definition of a limit gives $$\lim_{n\rightarrow+\infty}\sqrt{3+a_n}=\sqrt3.$$