0
$\begingroup$

Why does the following limit equals 2:

$$\lim_{x \to 0}\frac{2x^2}{\sin^2 x}=2$$

I can't find a trigonometric conversion to get that result.

  • 3
    Let $f(x)$ be your function. Then $f(x)= \frac{2}{\left( \frac{\sin x}{x}\right)^2}$.2012-05-17

3 Answers 3

4

Use the basic limit $$\lim_{x\to 0}\frac{\sin x}x=1$$ to derive this one. Then use some basic properties of limits. For example,

$$\lim_{x\to a}\frac1{f(x)}=\frac1{\lim\limits_{x\to a}f(x)}\;,$$

if the limit in the denominator is not zero.

  • 0
    But how do I convert my expression to that form? sin²x is dividing in this case2012-05-17
  • 1
    @JorgeZapata, only squating and inverting.2012-05-17
  • 0
    @Jorge: Gastón is correct, and I’ve added an extra hint as well.2012-05-17
  • 0
    But in this case the limit in the denominator is zero, isn't it?2012-05-17
  • 1
    @Jorge: Not if you pick the right $f$. HINT: $f(x)$ is **not** $\sin^2 x$.2012-05-17
  • 0
    Ok so I factor the 2 and divide my quadratic expressions by a product $2*\lim_{x \to 0}\frac{sinx}{x}*\frac{sinx}{x}$ But the result in this case would be 1 divided by the result 1/2??2012-05-17
  • 1
    @JorgeZapata If you know a value of some limit, in particular you know that limit exists and almost all aritmethic properties of limits you can use without problems. The properties that may you use are squaring and inverting. Use $f(x)=\frac{\sin x}{x}$ because is the only limit that you known but it is enought.2012-05-17
  • 1
    @Jorge: $$\lim_{x\to 0}\frac{2x^2}{\sin^2 x}=2\lim_{x\to 0}\frac1{\frac{\sin^2 x}{x^2}}=\dots\;?$$2012-05-17
  • 0
    Thanks @BrianM.Scott my problem was figure it out how to get that denominator. $$2\lim_{x \to0} \frac{x^2}{sin^2x}*\frac{\frac{1}{x^2}}{\frac{1}{x^2}}$$2012-05-17
0

$2\cdot \lim_{x\to 0} \dfrac{1}{\frac{\sin x}{x}} \cdot\dfrac{1}{\frac{\sin x}{x}} = 2(\frac{1}{1} \cdot \frac{1}{1}) = 2$

  • 0
    For some basic information about writing math at this site see e.g. [here](http://meta.math.stackexchange.com/questions/5020/), [here](http://meta.stackexchange.com/a/70559/155238), [here](http://meta.math.stackexchange.com/questions/1773/) and [here](http://math.stackexchange.com/editing-help#latex).2013-01-31
0

use Bernoulli's rule , wiki :http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule