Trig Limits.

**spoc21** · May 30th 2010, 10:10 PM

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

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

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

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

Thanks

**Prove It** · May 30th 2010, 10:23 PM

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

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

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

Thanks

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

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

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

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

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

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

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

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

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

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

**Prove It** · May 30th 2010, 10:26 PM

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

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

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

Thanks

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

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

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

$= \sin{(1)}$ .

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