Checagem Cuidadosa de Hipóteses

Help, I Broke L'Hospital's Rule!

What is wrong with the following applications of L'Hospital's Rule?

$$\lim_{x\to0}4x=\lim_{x\to0}4=4$$

$$\lim_{x\to1}\frac{x^2}{5x}=\lim_{x\to1}\frac{2x}{5}=\frac25$$

$$\lim_{x\to0}\frac{\cos x}{x}=\lim_{x\to0}\frac{-\sin x}1=0$$

$$\lim_{x\to0}\frac{\sin\sqrt{x}}{\sqrt{x}}=\lim_{x\to0}\frac{\frac1{2\sqrt x}\cos\sqrt{x}}{\frac1{2\sqrt{x}}}=\lim_{x\to0}\cos\sqrt{x}=1$$

Solutions

1. You can only apply L'Hospital's Rule on fractions!

2. Notice when you plug in $x=1$, you immediately get the limit to be $\frac15$.

3. Notice when you plug in $x=0$ you get $\frac10$, so the limit does not exist.

4. This was a tricky one, and I apologize ahead of time for being so mean by putting this one. For the limit to exist, the left-hand and right-hand limits exist. Although $\lim_{x\to0^+}\frac{\sin\sqrt x}{\sqrt x}=1$, $\frac{\sin\sqrt x}{\sqrt x}$ is not defined for $x<0$, so the limit actually doesn't exist.

These articles come with examples.

Caution 1

Caution 2

TODO: Make this interactive.