Trying to solve $\lim_{x\to1}{\frac {\cos(\frac {\pi x} {2})} {1-\sqrt x}}$ without L'Hospital

Question

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.

Answer 1

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 $$ \frac{\frac{\pi y}{2} }{\frac{y}{2}} = \pi $$

Answer 2

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 $\cos$.

Answer 3

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.

Answer 4

Let $f(x)=\cos\left(\frac{\pi x}2\right)$ then

$$\lim_{x\to1}{\frac {\cos(\frac {\pi x} {2})} {1-\sqrt x}}=2\lim_{x\to1}{\frac {f(x)-f(1)} {(1-\sqrt x)(1+\sqrt x)}}=-2\lim_{x\to1}\frac {f(x)-f(1)} {x-1}=-2f'(1)$$

Answer 5

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.