# How do you evaluate (1-cos2x)/(x^2) as x approaches 0?

##### 2 Answers
Write your answer here...
Start with a one sentence answer
Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Answer

Write a one sentence answer...

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

Describe your changes (optional) 200

5
Jim H Share
Mar 7, 2018

#### Answer:

Use $\cos 2 x = 1 - 2 {\sin}^{2} x$ and ${\lim}_{x \rightarrow 0} \sin \frac{x}{x} = 1$

#### Explanation:

${\lim}_{x \rightarrow 0} \frac{1 - \cos 2 x}{x} ^ 2 = {\lim}_{x \rightarrow 0} \frac{1 - \left(1 - 2 {\sin}^{2} x\right)}{x} ^ 2$

$= {\lim}_{x \rightarrow 0} \frac{2 {\sin}^{2} x}{x} ^ 2$

$= 2 {\left({\lim}_{x \rightarrow 0} \frac{\sin x}{x}\right)}^{2}$

$= 2 {\left(1\right)}^{2} = 2$

Was this helpful? Let the contributor know!
Write your answer here...
Start with a one sentence answer
Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Answer

Write a one sentence answer...

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

Describe your changes (optional) 200

3
Mar 7, 2018

#### Answer:

${\lim}_{x \setminus \rightarrow 0} {\frac{1 - \cos 2 x}{x}}^{2}$ is $2$.

#### Explanation:

If we try to substitute 0 into ${\lim}_{x \setminus \rightarrow 0} {\frac{1 - \cos 2 x}{x}}^{2}$, we end up with $\setminus \frac{0}{0}$. This introduces what is called an indeterminate form (there is another, infinity over infinity, but we don't have to worry about that one for this problem.)

When direct substitution yields an indeterminate form, we can use L'Hôpital's rule: ${\lim}_{x \setminus \rightarrow a} \frac{f \left(x\right)}{g \left(x\right)} = {\lim}_{x \setminus \rightarrow a} \frac{f ' \left(x\right)}{g ' \left(x\right)}$

Note that this is not the same as the Product Rule; rather, it means that the limit of the derivative of $f \left(x\right)$ over the derivative of $g \left(x\right)$ will be the same as the limit of $f \left(x\right)$ over $g \left(x\right)$.

Let's try this:
${\lim}_{x \setminus \rightarrow 0} {\frac{1 - \cos 2 x}{x}}^{2} = {\lim}_{x \setminus \rightarrow 0} \frac{2 \sin 2 x}{2} x$.

What happens when we substitute 0 back into this?
$\frac{2 \sin 2 \left(0\right)}{2 \left(0\right)} = \setminus \frac{2 \left(0\right)}{0}$

We end up with $\setminus \frac{0}{0}$ again, but L'Hôpital's rule says we can do this as many times as necessary, so if we apply it again:
${\lim}_{x \setminus \rightarrow 0} \frac{2 \sin 2 x}{2} x = {\lim}_{x \setminus \rightarrow 0} \frac{4 \cos 2 x}{2}$.

When we substitute 0 into this, we get
$\setminus \frac{4 \cos 2 \left(0\right)}{2} = \setminus \frac{4 \left(1\right)}{2}$

This gives if $\setminus \frac{4}{2}$, which reduces to $2$.

Thus, we know that ${\lim}_{x \setminus \rightarrow 0} {\frac{1 - \cos 2 x}{x}}^{2}$ is $2$.

For more on L’Hôpital's rule, I encourage you to check out this link .

Was this helpful? Let the contributor know!
##### Just asked! See more
• 16 minutes ago
• 18 minutes ago
• 18 minutes ago
• 22 minutes ago
• 4 minutes ago
• 4 minutes ago
• 7 minutes ago
• 11 minutes ago
• 14 minutes ago
• 14 minutes ago
• 16 minutes ago
• 18 minutes ago
• 18 minutes ago
• 22 minutes ago