Find the values of the following limits: 1) Lim (3-x)/(2+x) as x->2, 2) Lim (x2-9)/(x-3) as x->3, 3) Lim (x3-8)/(x2-4) as x->2, 4) Lim (sqrt(1+x2)-sqrt(1-x))/x2 as x->0, 5) Lim [(1-x2)/(1+x)] as x->+1

$$\frac{1}{4}$$

The value of the limit is $\boxed{\frac{1}{4}}$.

2) $\lim_{x \to 3} \frac{x^2-9}{x-3}$

$$3+3 = 6$$

The value of the limit is $\boxed{6}$.

3) $\lim_{x \to 2} \frac{x^3-8}{x^2-4}$

$$\frac{2^2+2(2)+4}{2+2} = \frac{4+4+4}{4} = \frac{12}{4} = 3$$

The value of the limit is $\boxed{3}$.

4) $\lim_{x \to 0} \frac{\sqrt{1+x^2}-\sqrt{1-x}}{x^2}$

$$\lim_{x \to 0} \frac{x+1}{x(\sqrt{1+x^2}+\sqrt{1-x})}$$

Upon re-evaluating, there was a mistake in the previous step. The $x$ in the denominator is $x^2$, so after canceling one $x$, there is still an $x$ left in the denominator. Let's re-do from Step 3.

Corrected Step 3: Simplify the numerator.

$$\lim_{x \to 0} \frac{1+x^2-1+x}{x^2(\sqrt{1+x^2}+\sqrt{1-x})} = \lim_{x \to 0} \frac{x^2+x}{x^2(\sqrt{1+x^2}+\sqrt{1-x})}$$

Corrected Step 4: Factor out $x$ from the numerator.

$$\lim_{x \to 0} \frac{x(x+1)}{x^2(\sqrt{1+x^2}+\sqrt{1-x})}$$

Corrected Step 5: Cancel out a common factor of $x$.

$$\lim_{x \to 0} \frac{x+1}{x(\sqrt{1+x^2}+\sqrt{1-x})}$$

This still results in $\frac{1}{0}$ if we substitute $x=0$. Let's check the original problem again.
The problem is $\lim_{x \to 0} \frac{\sqrt{1+x^2}-\sqrt{1-x}}{x^2}$.
Numerator: $(1+x^2) - (1-x) = x^2+x$.
So, $\frac{x^2+x}{x^2(\sqrt{1+x^2}+\sqrt{1-x})} = \frac{x(x+1)}{x^2(\sqrt{1+x^2}+\sqrt{1-x})} = \frac{x+1}{x(\sqrt{1+x^2}+\sqrt{1-x})}$.
This limit does not exist because the denominator approaches 0 while the numerator approaches 1.
Let's re-examine the problem statement or my interpretation.
The problem is $\lim_{x \to 0} \frac{\sqrt{1+x^2}-\sqrt{1-x}}{x^2}$.
If the problem was $\lim_{x \to 0} \frac{\sqrt{1+x^2}-\sqrt{1-x}}{x}$, then after multiplying by conjugate, we would get $\frac{x^2+x}{x(\sqrt{1+x^2}+\sqrt{1-x})} = \frac{x(x+1)}{x(\sqrt{1+x^2}+\sqrt{1-x})} = \frac{x+1}{\sqrt{1+x^2}+\sqrt{1-x}}$.
Substituting $x=0$ would give $\frac{1}{\sqrt{1}+\sqrt{1}} = \frac{1}{2}$.
However, the denominator is $x^2$.

Let's re-check the algebra carefully.

$$\lim_{x \to 0} \frac{\sqrt{1+x^2}-\sqrt{1-x}}{x^2}$$

Multiply by conjugate:

$$\lim_{x \to 0} \frac{\sqrt{1+x^2}-\sqrt{1-x}}{x^2} \cdot \frac{\sqrt{1+x^2}+\sqrt{1-x}}{\sqrt{1+x^2}+\sqrt{1-x}}$$ $$= \lim_{x \to 0} \frac{(1+x^2)-(1-x)}{x^2(\sqrt{1+x^2}+\sqrt{1-x})}$$ $$= \lim_{x \to 0} \frac{1+x^2-1+x}{x^2(\sqrt{1+x^2}+\sqrt{1-x})}$$ $$= \lim_{x \to 0} \frac{x^2+x}{x^2(\sqrt{1+x^2}+\sqrt{1-x})}$$ $$= \lim_{x \to 0} \frac{x(x+1)}{x^2(\sqrt{1+x^2}+\sqrt{1-x})}$$ $$= \lim_{x \to 0} \frac{x+1}{x(\sqrt{1+x^2}+\sqrt{1-x})}$$

As $x \to 0$, the numerator approaches $0+1=1$.
As $x \to 0$, the denominator approaches $0 \cdot (\sqrt{1+0^2}+\sqrt{1-0}) = 0 \cdot (1+1) = 0$.
Since the numerator approaches a non-zero number and the denominator approaches zero, the limit does not exist.
More specifically, if $x \to 0^+$, the denominator is $x \cdot (\text{positive number}) \to 0^+$, so the limit is $+\infty$.
If $x \to 0^-$, the denominator is $x \cdot (\text{positive number}) \to 0^-$, so the limit is $-\infty$.
Since the left and right limits are different, the limit does not exist.

Let's consider if L'Hopital's Rule is applicable.
The form is $\frac{0}{0}$.
Let $f(x) = \sqrt{1+x^2}-\sqrt{1-x}$ and $g(x) = x^2$.
$f'(x) = \frac{1}{2\sqrt{1+x^2}}(2x) - \frac{1}{2\sqrt{1-x}}(-1) = \frac{x}{\sqrt{1+x^2}} + \frac{1}{2\sqrt{1-x}}$
$g'(x) = 2x$

$$\lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \frac{\frac{x}{\sqrt{1+x^2}} + \frac{1}{2\sqrt{1-x}}}{2x}$$ $$= \lim_{x \to 0} \left( \frac{x}{2x\sqrt{1+x^2}} + \frac{1}{4x\sqrt{1-x}} \right)$$ $$= \lim_{x \to 0} \left( \frac{1}{2\sqrt{1+x^2}} + \frac{1}{4x\sqrt{1-x}} \right)$$

As $x \to 0$, the first term approaches $\frac{1}{2\sqrt{1}} = \frac{1}{2}$.
The second term approaches $\frac{1}{0}$, which means it does not exist.
So, L'Hopital's Rule also indicates the limit does not exist.

It's possible the problem intended a different denominator or a different numerator. Given the image, $x^2$ is clearly the denominator.
I will state that the limit does not exist.

As $x \to 0$, the numerator approaches $0+1=1$. The denominator approaches $0 \cdot (\sqrt{1+0^2}+\sqrt{1-0}) = 0 \cdot (1+1) = 0$.
Since the numerator approaches a non-zero value and the denominator approaches zero, the limit does not exist.
The value of the limit is $\boxed{\text{Does Not Exist}}$.

5) $\lim_{x \to 1} \frac{1-x^2}{1+x}$

$$1-1 = 0$$

The value of the limit is $\boxed{0}$.