# Thread: trig and limits problem

1. ## trig and limits problem

hi,

using sin(θ)/θ x->0 = 1 explain how
a.) lim (sin2x)/5x = 2/5 as x ->0
b.)lim (cos h -1)/h = 0 as h->0 using the half angle formula cos h = 1 - 2 sin^ * (h/2)

since a. is an identity do does the substitution rule apply?

in b. i can set it up:
lim -((2sin^2 )*(h/2)/h)

im not sure what to do or how to solve this limit. my algebra is a little weak so can you explain it to me in simple terms?? any response would be great!

2. ## Re: trig and limits problem

Originally Posted by kingargon
hi,

using sin(θ)/θ x->0 = 1 explain how
a.) lim (sin2x)/5x = 2/5 as x ->0
b.)lim (cos h -1)/h = 0 as h->0 using the half angle formula cos h = 1 - 2 sin^ * (h/2)

since a. is an identity do does the substitution rule apply?

in b. i can set it up:
lim -((2sin^2 )*(h/2)/h)

im not sure what to do or how to solve this limit. my algebra is a little weak so can you explain it to me in simple terms?? any response would be great!
a)

substitute $2x=t \Rightarrow 5x=\frac{5t}{2}$

also : $x \rightarrow 0 \Rightarrow t \rightarrow 0$

Hence :

$L=\displaystile \lim_{t \to 0} \frac{\sin t}{\frac{5t}{2}}=\displaystile \lim_{t \to 0}\left (\frac{2}{5} \cdot \frac{\sin t}{t}\right)=\frac{2}{5} \cdot \displaystile \lim_{t \to 0} \frac{\sin t}{t} =\frac{2}{5}$

b)

Hint :

$\displaystile \lim_{t \to 0} \frac{\sin^2 t}{t}=\displaystile \lim_{t \to 0} \sin t \cdot \displaystile \lim_{t \to 0} \frac{\sin t}{t}$

3. ## Re: trig and limits problem

great answer! thank you for the quick response.

i dont understand some of your steps can you explain how u got the substitution values and your arithmitic to find the limit?

4. ## Re: trig and limits problem

Originally Posted by kingargon
great answer! thank you for the quick response.
i dont understand some of your steps can you explain how u got the substitution values and your arithmitic to find the limit?
$\frac{\sin(2x)}{5x}=\left(\frac{2}{5}\right)\left( \frac{\sin(2x)}{2x}\right)$
Now if $x\to 0$ then also $2x\to 0$