# Why is $\lim_{x\to0}\frac{2-\sqrt{4-x}}{x} = \frac{1}{4}$?

$$\lim_{x\to0} \frac{2-\sqrt{4-x}}{x} = \frac{1}{4}$$

Whether completing-the-square nor adding-zero helped me converting this equation into something useful. In the end, there is always one $x$ that 'zeros' my terms. I'm sure I've tried every possible trick I know. Still, there must be some technique left. What did I miss?

• Notice that $2^2-4=0$ ;) – Mann Dec 31 '14 at 11:22
• Multiply both numerator and denominator by $2+\sqrt{(4-x)}$ – Volodymyr Fomenko Dec 31 '14 at 11:23
• @user1511417 It does – Alice Ryhl Dec 31 '14 at 11:24
• How many more answers saying the same thing do we need? – Alice Ryhl Dec 31 '14 at 11:27
• Just a note: $2+\sqrt{4-x}$ is called the conjugate of $2-\sqrt{4-x}$ (so just the - changed into a +). This multiplication results in a product of the form $(a+b)*(a-b)=a^2-b^2$, which is a lot easier to deal with. – Dasherman Dec 31 '14 at 11:28

$$\frac{2-\sqrt{4-x}}{x}=\frac{2-\sqrt{4-x}}{x}\frac{2+\sqrt{4-x}}{2+\sqrt{4-x}}=\frac{2^2-(4-x)}{x\cdot(2+\sqrt{4-x})}=\frac{1}{2+\sqrt{4-x}}$$

• Hm? 2² - 4 - 0 is 0 and 0*(2+....) is also 0 – user1511417 Dec 31 '14 at 11:28
• @user1511417 The 4s cancel on the numerator to 0, and then the $x$ on top and bottom cancel – Alice Ryhl Dec 31 '14 at 11:31
• Ah I see your point. .. ok – user1511417 Dec 31 '14 at 11:33

Let $y>0$ be such that $y^2=4-x$, then $$\lim_{x\to 0}\frac{2-\sqrt{4-x}}x=\lim_{y\to 2}\frac{2-y}{4-y^2}=\lim_{y\to 2}\frac{2-y}{(2-y)(2+y)}=\lim_{y\to 2}\frac1{2+y}=\frac14$$

• That's an interesting method. – user1511417 Dec 31 '14 at 11:35
• Gosh! This is ingenious! – user1511417 Dec 31 '14 at 11:42

You can use $(a+b)(a-b) = a^2-b^2$ and get

$$\lim_{x\to 0} \frac{4-4+x}{x(2+\sqrt{4-x})} = \lim_{x \to 0} \frac{1}{2+\sqrt{4-x}}.$$

Hint:

Do you see why this limit is the same of

$$\lim \limits_{x \rightarrow 0}\frac{1}{2+ \sqrt{4-x}}?$$

• No, I am sorry. – user1511417 Dec 31 '14 at 11:26
• See Volodymyr's comment. – Alex Silva Dec 31 '14 at 11:27