using $\epsilon$ and $\delta$ prove $\lim_{x\to 1}\frac{x+3}{1+\sqrt x}=2$

Question

$$\lim_{x\to 1}\frac{x+3}{1+\sqrt x}=2$$

i got

$$|f(x)-2|<|x-1||\sqrt x -1||\frac{1}{(1+\sqrt x)^2}|$$

$\to \delta=1,\epsilon=\sqrt 2-1$

Is that true ??

Answer 1

You cannot just set the value of epsilon. You have to, for an arbitrary value of $\epsilon$, find such a value of $\delta$ that $|f(x)-2|<\epsilon$ for every $x$ for which $|x-1|<\delta$.

My advice:

Try to remove $x$ from the estimates of $|f(x) - 2|$. For example, you already have

$$|f(x)-2|<|x-1|\cdot C(x)$$

You already know that $|x-1|$ is smaller than $\delta$.