Calculate $ \lim_{x \to 4} \frac{3 -\sqrt{5 -x}}{1 -\sqrt{5 -x}} $ How evaluate the following limit?
$$ \lim_{x \to 4} \frac{3 -\sqrt{5 -x}}{1 -\sqrt{5 -x}} $$
I cannot apply L'Hopital because $ f(x) =  3 -\sqrt{5 -x} \neq 0 $ at $x = 5$
 A: I think the problem is $$\lim_{x\to4}\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}$$
HINT:
$$\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}=\frac{3^2-(5+x)}{1^2-(5-x)}\cdot\frac{1+\sqrt{5-x}}{3+\sqrt{5+x}}=-\frac{1+\sqrt{5-x}}{3+\sqrt{5+x}}$$ if $x\ne4$
Now here $x\to4\implies x\ne4$
A: As you've surmised, the denominator tends to zero while the numerator does not; hence, the expression blows up in absolute value.
On the other hand, if $x < 4$, the denominator is negative and the overall expression is positive; but if $x > 4$, the denominator is positive. Hence, as $x$ approaches from the right, the expression tends to $\infty$; but from the left, it tends to $-\infty$. Hence, the limit does not exist.
This can be confirmed by looking at the graph.
A: Let $\sqrt{5+x}=3+h$
Then $5+x=9+6h+h^2$
Let $\sqrt{5-x}=1+k$
Then $5-x=1+2k+k^2$
Adding gives: $5+x+5-x=9+6h+h^2+1+2k+k^2$
$0=6h+h^2+2k+k^2$
When $x=4$, $\sqrt{5+x}=\sqrt 9 =3 \Rightarrow 3+h=3 \Rightarrow h=0$
Similarly $k=0$
Required expression is $\frac {3-\sqrt{5+x}}{1-\sqrt{5-x}}=\frac h k$
$6h=-2k-k^2-h^2$
$\frac h k= -\frac 2 6 -\frac k6-\frac {h^2} {6k}$
As $h$ and $k$ tend to 0, $\frac h k$ tends to $-\frac 1 3$
A: Let $x=5\cos4y$  where $0\le4y\le\pi$
$x\to4\implies\cos4y\to\dfrac45$
But as $\cos4y=2\cos^22y-1, 2y\to\arccos\dfrac3{\sqrt{10}}=\arcsin\dfrac1{\sqrt{10}}$
$$F=\lim_{x\to4}\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}=\lim_{2y\to\arcsin\frac1{\sqrt{10}}}\dfrac{\dfrac3{\sqrt{10}}-\cos2y}{\dfrac1{\sqrt{10}}-\sin2y}$$
If we set $\arccos\dfrac3{\sqrt{10}}=\arcsin\dfrac1{\sqrt{10}}=2A\implies\cos2A=?,\sin2A=?,\tan2A=\dfrac13$
$$F=\lim_{2y\to2A}\dfrac{\cos2A-\cos2y}{\sin2A-\sin2y}=\lim_{y\to A}\dfrac{-2\sin(A-y)\sin(A+y)}{2\sin(A-y)\cos(A+y)}=-\tan(A+A)=?$$
