# Thread: Trig Limit Help neded

1. ## Trig Limit Help neded

Hello!

I've been having trouble solving these 2 trig limits, any help would be appreciated.

find the limit as x -> 0
1) sin7x / sin4x

2) [sin (cos x)] / sec x

I need to re-arrange both these equations so that 0 can be subbed into x and not get 0 /0. For 1) i multiplied the equation by 7x over 7x and re-arranged it to (sin7x) / 7x * (7x) / sin4x. The sin7x / 7x becomes 1 because of the trig limit properties, but what do I do with 7x / sin4x?

I don't even know how to start 2

For those of you that don't know the 2 trig limit properties of the top of your head, here they are:
limit as x -> 0 of sin x / x = 1
limit as x -> 0 of cos x - 1 / x = 0

Thanks for any help
-Jackson

2. $\frac{{\sin \left( {7x} \right)}}{{\sin \left( {4x} \right)}} = \frac{{\left( {1/x} \right)}}{{\left( {1/x} \right)}}\left( {\frac{{\sin \left( {7x} \right)}}{{\sin \left( {4x} \right)}}} \right) = \frac{{\frac{{7\sin (7x)}}{{7x}}}}{{\frac{{4\sin (4x)}}{{4x}}}}$

P.S. #2 is just $\sin (1)$. WHY?

3. Originally Posted by TheJacksonater
...
2) [sin (cos x)] / sec x

...
$\frac{\sin(\cos(x))}{\sec(x)}= \frac{\sin(\cos(x))}{\frac1{\cos(x)}}$

And therefore

$\lim_{x \rightarrow 0}\left(\dfrac{\sin(\cos(x))}{\dfrac1{\cos(x)}}\ri ght) = \sin(1)$

4. Thanks, wow those were easier then I thought now that I see the solutions. Makes me feel pretty stupid