1
$\begingroup$

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

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?

$\endgroup$
  • $\begingroup$ Notice that $2^2-4=0$ ;) $\endgroup$ – Mann Dec 31 '14 at 11:22
  • 2
    $\begingroup$ Multiply both numerator and denominator by $2+\sqrt{(4-x)}$ $\endgroup$ – Volodymyr Fomenko Dec 31 '14 at 11:23
  • 1
    $\begingroup$ @user1511417 It does $\endgroup$ – Alice Ryhl Dec 31 '14 at 11:24
  • 1
    $\begingroup$ How many more answers saying the same thing do we need? $\endgroup$ – Alice Ryhl Dec 31 '14 at 11:27
  • 1
    $\begingroup$ 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. $\endgroup$ – Dasherman Dec 31 '14 at 11:28
9
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Hm? 2² - 4 - 0 is 0 and 0*(2+....) is also 0 $\endgroup$ – user1511417 Dec 31 '14 at 11:28
  • $\begingroup$ @user1511417 The 4s cancel on the numerator to 0, and then the $x$ on top and bottom cancel $\endgroup$ – Alice Ryhl Dec 31 '14 at 11:31
  • $\begingroup$ Ah I see your point. .. ok $\endgroup$ – user1511417 Dec 31 '14 at 11:33
5
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ That's an interesting method. $\endgroup$ – user1511417 Dec 31 '14 at 11:35
  • $\begingroup$ Gosh! This is ingenious! $\endgroup$ – user1511417 Dec 31 '14 at 11:42
2
$\begingroup$

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

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

$\endgroup$
2
$\begingroup$

Hint:

Do you see why this limit is the same of

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.