stuck on trig. limit

[letsgetaway]

\(\displaystyle \[
\begin{array}{l}
\lim 1 - \sec ^2 (x)2x/x^2 \\
x \to 0 \\
\end{array}
\]\)

My work so far...

\(\displaystyle \[
- \tan ^2 (x)2x/x^2 = (2)[ - \sin ^2 (x)/\cos ^2 (x)]
\]\)

I'm not sure how to approach this problem after the step I made. The book's answer is "-4" but with my approach I am only able to get "-2".

 

[mark07]

Do you mean

\(\displaystyle \L
\lim_{x \to 0} \frac{ \left( 1 - \sec ^2 (x) \right) 2x}{x^2} ?\)

If so, you have

\(\displaystyle \L
\lim_{x \to 0} \frac{ 2 \left( 1 - \sec ^2 (x) \right) }{x}
= \lim_{x \to 0} \frac{ -2 \tan^2 (x) }{x}\)

But then if you use L'Hospital's rule, limit turns out to be 0, not -4.

Please type the question again and we'll see.

 

[letsgetaway]

I am going to try doing it on my own with what you've written. I still haven't gotten it though. My class doesn't use L'Hospital's rule. I heard of it, but I don't know how to apply it.

Update....

I can't believe this. I wrote the problem incorrectly from the book. No wonder I was having so much trouble.

Thanks, Mark.