Simple limit problems I can't figure out

[cjucalc]

I'm studying for a test, and can't figure out (1) limx->0 tan5x/sec3x, (2) lim x->0 1-cos2x/x, and (3) 1-cosx/sinx. (1) I'm at sin5x/cos5x*cos3x, (2) I think it's 2 because 2*(1-cos2x/2x) goes to 2, and (3) I keep getting x/sinx*1-cosx/2x which goes to 1, but the answer's supposedly 0. Thanks!

 

[soroban]

Hello, cjucalc!!

We are expected to know these two theorems:

. . \(\displaystyle \displaystyle\lim_{ \theta\to0}\frac{\sin\theta}{\theta} \:=\:1\)

. . \(\displaystyle \displaystyle \lim_{\theta\to0}\frac{1-\cos\theta}{\theta} \:=\:0\)

\(\displaystyle \displaystyle(1)\;\lim_{x\to0}\frac{\tan5x}{\sec3x}\)

Did you substitute \(\displaystyle x=0\,?\)

\(\displaystyle \displaystyle(2)\;\lim_{x\to0}\frac{1-\cos2x}{x}\)

Multiply by \(\displaystyle \frac{2}{2}\!:\)

\(\displaystyle \displaystyle \lim_{x\to0}\;\frac{2}{2}\!\cdot\!\frac{1-\cos2x}{x} \;=\;2\cdot\underbrace{\lim_{x\to0}\frac{1-\cos2x}{2x}}_{\text{This is 0}} \;=\;2\cdot 0 \;=\;0\)

\(\displaystyle \displaystyle(3)\;\lim_{x\to0}\frac{1-\cos x}{\sin x}\)

Divide numerator and denominator by \(\displaystyle x\!:\)

. . \(\displaystyle \displaystyle\lim_{x\to0}\frac{\frac{1-\cos x}{x}}{\frac{\sin x}{x}} \;=\;\frac{\overbrace{\lim\:\frac{1-\cos x}{x}}^{\text{This is 0}}}{\underbrace{\lim\:\frac{\sin x}{x}}_{\text{This is 1}}} \;=\;\frac{0}{1} \;=\;0 \)