1. ## Trig Limits.

Hi, I'm having a lot of trouble in solving the following limit problems (all of the, as x --> 0):

1) $\displaystyle \frac{sin 7x}{sin 4x}$

2) $\displaystyle \frac{sin(cos x)}{sec x}$

I'm very confused with these two problems, and would greatly appreciate any help (I'm not allowed to differentiate the functions).

Thanks

2. Originally Posted by spoc21
Hi, I'm having a lot of trouble in solving the following limit problems (all of the, as x --> 0):

1) $\displaystyle \frac{sin 7x}{sin 4x}$

2) $\displaystyle \frac{sin(cos x)}{sec x}$

I'm very confused with these two problems, and would greatly appreciate any help (I'm not allowed to differentiate the functions).

Thanks
$\displaystyle \lim_{x \to 0}\frac{\sin{7x}}{\sin{4x}} = \lim_{x \to 0}\frac{\sin{(3x + 4x)}}{\sin{4x}}$

$\displaystyle = \lim_{x \to 0}\frac{\sin{3x}\cos{4x} + \cos{3x}\sin{4x}}{\sin{4x}}$

$\displaystyle = \lim_{x \to 0}\left(\frac{\sin{3x}\cos{4x}}{\sin{4x}} + \cos{3x}\right)$

$\displaystyle = \lim_{x \to 0}\left(\frac{\sin{3x}\cos{4x}}{\sin{4x}}\right) + \lim_{x \to 0}\,(\cos{3x})$

$\displaystyle = 1 + \lim_{x \to 0}\left(\frac{\sin{3x}\cos{4x}}{\sin{4x}}\right)$

$\displaystyle = 1 + \lim_{x \to 0}\left[\frac{(3\sin{x} - 4\sin^3{x})(1 - 2\sin^2{2x})}{2\sin{2x}\cos{2x}}\right]$

$\displaystyle = 1 + \lim_{x \to 0}\left[\frac{\sin{x}(3 - 4\sin^2{x})(1 - 2\sin^2{2x})}{4\sin{x}\cos{x}\cos{2x}}\right]$

$\displaystyle = 1 + \lim_{x \to 0}\left[\frac{(3 - 4\sin^2{x})(1 - 2\sin^2{2x})}{4\cos{x}\cos{2x}}\right]$

$\displaystyle = 1 + \frac{3}{4}$

$\displaystyle = \frac{7}{4}$.

3. Originally Posted by spoc21
Hi, I'm having a lot of trouble in solving the following limit problems (all of the, as x --> 0):

1) $\displaystyle \frac{sin 7x}{sin 4x}$

2) $\displaystyle \frac{sin(cos x)}{sec x}$

I'm very confused with these two problems, and would greatly appreciate any help (I'm not allowed to differentiate the functions).

Thanks
$\displaystyle \lim_{x \to 0}\frac{\sin{(\cos{x})}}{\sec{x}} = \lim_{x \to 0}\,[\cos{x}\sin{(\cos{x})}]$

$\displaystyle = \cos{(0)}\sin{[\cos{(0)}]}$

$\displaystyle = 1\cdot \sin{(1)}$

$\displaystyle = \sin{(1)}$.