trig limits: [tan(x+h) - tan(x)] / h, as h goes to zero

[Cuddles]

lim tan(x+h)-tan(x)
h->0 h

I got as far as (tan(x)+tan(h)(1-tan^2(x)))/(h(1+tan(x)tan(h)))

I'm not sure if that's right or I'm on the right track or not. My answer choices are
A. 1
B. sec(x)
C. sec(x)tan(x)
D. sec^2(h)
E. None of these

 

[o_O]

Re: trig limits

Using the double angle formula, you should get:

\(\displaystyle = \lim_{h \to 0}\frac{\frac{tan(x)+tan(h)}{1-tan(x)tan(h)} - tan(x)}{h}\)

My advice is to combine the numerator into one fraction:

\(\displaystyle \lim_{h \to 0} \left[\frac{1}{h}\left(\frac{tan(x)+tan(h)}{1-tan(x)tan(h)} - \frac{tan(x)\left(1 - tan(x)tan(h)\right)}{1 - tan(x)tan(h)}\right)\right]\)

\(\displaystyle \lim_{h \to 0} \frac{1}{h} \cdot \frac{tanx + tanh - tanx(1 - tanxtanh)}{1-tanxtanh}\)

etc. etc.

Key things that will be useful:

\(\displaystyle \lim_{h \to 0}\frac{tanh}{h} = \lim_{h \to 0}\frac{sinh}{h} \cdot \frac{1}{cosh} = 1 \cdot 1 = 1\)

\(\displaystyle \mbox{Trig identity:} \: 1 + tan^{2}(x) = \mbox{???}\)

 

[skeeter]

lim tan(x+h)-tan(x)
h->0 h

I got as far as (tan(x)+tan(h)(1-tan^2(x)))/(h(1+tan(x)tan(h)))

I'm not sure if that's right or I'm on the right track or not. My answer choices are
A. 1
B. sec(x)
C. sec(x)tan(x)
D. sec^2(h)
E. None of these

this is a "recognition" problem ... remember this?

\(\displaystyle f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}\)

now .. look at your limit ... what is f(x) and what is f'(x) (the limit) ?