1. ## limit trigo function

quite rusty...

Would anyone help with the following, please:

limc_pi/2[(X+tanC)/(1-XtanC)] =?

I don't understand why, in a book, they continue thus:

limc_pi/2[sec2C/(-Xsec2C] = -1/X

.

2. ## Re: limit trigo function

Your book is using L'Hôpital's rule, which essentially states that if given:

$\displaystyle L=\lim_{u\to c}\frac{f(u)}{g(u)}$

and:

$\displaystyle \lim_{u\to c}f(u)=\lim_{u\to c}g(u)=0$ or $\displaystyle \lim_{u\to c}f(u)=\lim_{u\to c}g(u)=\infty$

then:

$\displaystyle L=\lim_{u\to c}\frac{f'(u)}{g'(u)}$

Do you see that if you write the limit you are given as:

$\displaystyle -\lim_{c\to\frac{\pi}{2}}\frac{x+\tan(x)}{x\tan(c)-1}$

you have the second indeterminate form above $\displaystyle \frac{\infty}{\infty}$?

3. ## Re: limit trigo function

Thank you, Mark.
the
The indetermination was readily obvious to me.
My confusion relates to the transition into the sec^2 expressions.

Would you please comment on that?

Thanks.

4. ## Re: limit trigo function

They are a result of:

$\displaystyle \frac{d}{du}(\tan(u))=\sec^2(u)$.

5. ## Re: limit trigo function

Of course!

Thank you very much, Mark!