1. ## Trig Limits

I'm trying to evaluate this limit and i'm having some issues.

$\lim_{x\to 0} \frac{1-\sec^2(2x)}{x^2}$

I'm not 100% sure about my trig Identities but i think that
$1-\sec^2(2x)$ is equal to $2\tan^2(x)$
is that correct?

if so, i say that the equation is equal to $\frac{2\sin^2(x)}{2\cos^2(x)}(\frac{1}{x^2})$

I have a feeling this is wrong somehow knowing the answer is -4. Can someone help me?

thank you

2. Originally Posted by discobob
I'm trying to evaluate this limit and i'm having some issues.

$\lim_{x\to 0} \frac{1-\sec^2(2x)}{x^2}$

I'm not 100% sure about my trig Identities but i think that
$1-\sec^2(2x)$ is equal to $2\tan^2(x)$
is that correct?

if so, i say that the equation is equal to $\frac{2\sin^2(x)}{2\cos^2(x)}(\frac{1}{x^2})$

I have a feeling this is wrong somehow knowing the answer is -4. Can someone help me?

thank you
The identity is $sec^2(\theta)=1+tan^2(\theta)$.
for any well-defined $\theta$.
Now, Try to correct your mistake.

Hint: Try to get the minus sign as a common factor from the numerator.

3. ok, this sounds so easy but i'm not so sure on how to get . That should be pretty basic but nothing is coming to mind. I mean, i know you can factor out a -1 getting $-1(-1+\sec^2(2x))$. But that's not what you mean right?

4. $

lim_{x->0} ((1-sec˛2x)/(x˛))
= lim_{x->0}((cos˛2x-1)/(x˛))
= lim_{x->0}((-(1-cos˛2x)4)/((2x)˛))
=lim_{x->0}(((sin˛2x)(-4))/((2x)˛))
=-4

$

5. $

lim_{x->0} ((1-sec˛2x)/(x˛))
= lim_{x->0}((cos˛2x-1)/(x˛))
= lim_{x->0}((-(1-cos˛2x)4)/((2x)˛))
=lim_{x->0}(((sin˛2x)(-4))/((2x)˛))
=-4

$

6. lim_{x->0} ((1-sec˛2x)/(x˛))
= lim_{x->0}((cos˛2x-1)/(x˛))
= lim_{x->0}((-(1-cos˛2x)4)/((2x)˛))
=lim_{x->0}(((sin˛2x)(-4))/((2x)˛))
=-4

7. i forgot the cos^2(2x) in the denominator.nevertheless that would vanish as cos 0=1 after putting the limits.