Trying to solve $\lim_{x\to1}{\frac {\cos(\frac {\pi x} {2})} {1-\sqrt x}}$ without L'Hospital I'm trying to solve the following limit without L'Hopital
$$\lim_{x\to1}{\frac {\cos(\frac {\pi x} {2})} {1-\sqrt x}}$$
But I have no clue how to solve this. Rationalization doesn't get me anywhere with the denominator, and I don't see anything to do with numerator.
 A: Define $f(x)=\cos \left ( \frac{\pi x}{2} \right ) \left ( 1 + \sqrt{x} \right )$. By rationalization and factoring a minus sign, you have
$$-\frac{f(x)}{x-1}$$
Since $f(1)=0$ this can be rewritten as
$$-\frac{f(x)-f(1)}{x-1}$$
So the limit of it as $x \to 1$ is $-f'(1)$, by the definition of the derivative. Calculating $f'(1)$ by a limit process seems somewhat difficult; you would need to replicate the argument for the product rule and the argument for the derivative of $\cos$.
A: Let $f(x)=\cos\left(\frac{\pi x}2\right)$ then
$$\lim_{x\to1}{\frac {\cos(\frac {\pi x} {2})} {1-\sqrt x}}=2\lim_{x\to1}{\frac {f(x)-f(1)} {(1-\sqrt x)(1+\sqrt x)}}=-2\lim_{x\to1}\frac {f(x)-f(1)} {x-1}=-2f'(1)$$
A: Hint: Use the substitution $y=x-1$, your limit would be then equal to 
$$
\lim_{y\to0}\dfrac{\cos(\tfrac\pi2(y+1))}{1-\sqrt{y+1}},
$$
after using the angle-addition formula and rationalizing the denominator you'll get :
$$
\lim_{y\to0}\dfrac{\sin(\tfrac\pi2y)(1+\sqrt{y+1})}{y},
$$
which is simpler to deal with.
A: A brute-force L'Hospital avoider: Substitute $h=1-\sqrt x$, and you get
$$ \lim_{h \to 0}\frac{\cos(\frac\pi2  (1-h)^2)}{h} $$
which is the derivative of $\cos(\frac\pi2(1-x)^2)$ at $x=0$. Differentiate symbolically and evaluate.
A: The straightforward way is to let $y = 1-x$.  Then 
$$\cos\frac{\pi x}{2} = \cos \left( \frac{\pi}{2} (1-y) \right) = \sin \frac{\pi y}{2} 
= \frac{\pi y}{2} + O(y^3)$$
$$ 1-\sqrt{x} = 1 - \sqrt{1-y} = \frac{y}{2} + O(y^2)
$$
So for sufficiently small $y$, the function gets arbitrarily close to 
$$ \frac{\frac{\pi y}{2} }{\frac{y}{2}} = \pi
$$
