Indeterminate limits involving square roots So, I've been for the past $2$ days trying to solve these two limits:
$$\lim_{x \to 4} \frac{3 - \sqrt{5 + x}}{1 - \sqrt{5 - x}}$$
$$\lim_{x \to 2} \frac{x^2 + 3x - 10}{3x^2 - 5x - 2}.$$
My problem with the first one is that multiplying both the denominator and numerator by either $3 + \sqrt{5 + x}$ or $1 + \sqrt{5 - x}$ simply leads me nowhere.
Regarding the second limit, I can't find $(x + a)$ that divides both the numerator and denominator, which would eliminate the indetermination.
 A: For question $1$, you were on the right lines when you tried to multiply the numerator and denominator by $1+\sqrt{5-x}$:
$$
\lim_{x\to4}\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}=\lim_{x\to4}\frac{\left(3-\sqrt{5+x}\right)\left(1+\sqrt{5-x}\right)}{x-4}
$$
From here, you might try to expand the numerator, which still leaves us with an indeterminate form. Better is to use the product law for limits:
$$
\lim_{x\to4}\frac{\left(3-\sqrt{5+x}\right)\left(1+\sqrt{5-x}\right)}{x-4}=\lim_{x\to4}\left(1+\sqrt{5-x}\right) \cdot \lim_{x\to4}\frac{3-\sqrt{5+x}}{x-4} \, .
$$
It's difficult to spot that this is the correct approach, but it helps to notice that $1+\sqrt{5-x}$ does not vanish when you plug in $x=4$. We can't write the limit as
$$
\lim_{x\to4}\left(3-\sqrt{5+x}\right) \cdot \lim_{x\to4}\frac{1+\sqrt{5-x}}{x-4}
$$
because the second limit does not exist.*
To compute
$$
\lim_{x\to4}\frac{3-\sqrt{5+x}}{x-4} \, ,
$$
multiply the numerator and denominator by $3+\sqrt{5+x}$.
Question $2$ is much easier. Notice that $x^2+3x-10=(x+5)(x-2)$. Also,
\begin{align}
3x^2 - 5x -2 &= 3x^2 - 6x + x - 2 \\
&= 3x(x-2)+(x-2) \\
&= (3x+1)(x-2) \, .
\end{align}
(Here, I've used a factorisation trick that works nicely when the leading coefficient is not equal to $1$. You can also complete the square to find the factors.) Once you have cancelled the common factor, the computation becomes trivial.

*The product law for limits states that if $\lim_{x \to a}f(x)$ and $\lim_{x\to a}g(x)$ exist, then $\lim_{x\to a}f(x)g(x)$ exists, and
$$
\lim_{x\to a}f(x)g(x)=\lim_{x\to a}f(x) \cdot \lim_{x\to a}g(x) \, .
$$
However, there might be cases where $\lim_{x\to a}f(x)g(x)$ exists, despite the fact that $\lim_{x \to a}f(x)$ or $\lim_{x \to a}g(x)$ might not exist. This is why despite the fact that the following limit exists
$$
\lim_{x\to4}\frac{\left(3-\sqrt{5+x}\right)\left(1+\sqrt{5-x}\right)}{x-4} \, ,
$$
the individual limits
$$
\lim_{x\to4}\left(3-\sqrt{5+x}\right)
$$
and
$$
\lim_{x\to4}\frac{1+\sqrt{5-x}}{x-4} \tag{*}\label{*}
$$
need not exist.
