This is the limit: $$ \lim_{x\to 0} \frac{\sqrt{1-x}}{\sqrt{1+x}} = 1$$
I want to verify by the $ \epsilon - \delta $ definition that the limit of this function is 1.
My process: Let $\epsilon > 0$ and we try to find a $\delta$ such that, if $ 0<\left | x - a \right | < \delta $ then $ \left |f(x)-L\right | < \epsilon$. In this case: $$ f(x) = \frac{\sqrt{1-x}}{\sqrt{1+x}} \qquad a = 0 \qquad L = 1 $$ Then: $$\left |f(x)-L\right | < \epsilon \Rightarrow \left |\frac{\sqrt{1-x}}{\sqrt{1+x}} - 1 \right | < \epsilon$$ I transformated the function by multiplying by the conjugate of the denominator in the numerator and in the denominator. Then, I eliminated the denominator, so the new function is bigger. $$ \left |\frac{1-x}{\sqrt{1-x^2}} - 1 \right | < \left |1- x - 1 \right |$$ Finally, I changed the negative sign, because the absolute value must be the same for positive x. $$\left |-x\right | = \left | x \right | < \delta = \epsilon$$
Is it correct?
I am sorry, your proof lacks two things:
First and most important, logic connections. How are the statements linked?
Second, you should explain what is the relation between the second and third lines.