Trig Limits

**discobob** · January 24th 2010, 08:31 AM

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

**General** · January 24th 2010, 08:39 AM

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.

**discobob** · January 24th 2010, 09:02 AM

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?

**Pulock2009** · January 24th 2010, 09:42 AM

$ lim_{x->0} ((1-sec²2x)/(x²)) = lim_{x->0}((cos²2x-1)/(x²)) = lim_{x->0}((-(1-cos²2x)4)/((2x)²)) =lim_{x->0}(((sin²2x)(-4))/((2x)²)) =-4 $

**Pulock2009** · January 24th 2010, 09:51 AM

$ lim_{x->0} ((1-sec²2x)/(x²)) = lim_{x->0}((cos²2x-1)/(x²)) = lim_{x->0}((-(1-cos²2x)4)/((2x)²)) =lim_{x->0}(((sin²2x)(-4))/((2x)²)) =-4 $

**Pulock2009** · January 24th 2010, 09:54 AM

lim_{x->0} ((1-sec²2x)/(x²))
= lim_{x->0}((cos²2x-1)/(x²))
= lim_{x->0}((-(1-cos²2x)4)/((2x)²))
=lim_{x->0}(((sin²2x)(-4))/((2x)²))
=-4

**Pulock2009** · January 24th 2010, 10:00 AM

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

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