1. ## trigonometric limits

Hey guys im new here, but I have a test today and I need to understand trig limits better.

lim of x approaching 0
sin(4x)
sin(6x)

the hint that was given was "use the identity Lim x approaching 0
sinx = 1"
x

2. oh and another question, which is on the def. of a derivitive, i just cannot remember the equation of a tangent line/equation of the normal line.

3. Multiply by $\frac{x}{x}$, so that you would get:

$\frac{\sin{4x}}{x} \cdot \frac{x}{\sin{3x}}$

What's left is simple manipulation to get your limit into something like sin(u)/u

4. Originally Posted by fortplainman
oh and another question, which is on the def. of a derivitive, i just cannot remember the equation of a tangent line/equation of the normal line.
The derivative is the slope of the tangent line. The equation of a line in point slope form is given by:

$y - y_1 = m(x-x_1)$

Where $(x_1, y_1)$ is the point in question and m is the slope. The equation of the normal line to this line contains $(x_1, y_1)$, but its slope is the negative reciprocal of the original line.

For instance, if line K has a slope of 4, then the line that is perpendicular/normal to line k will have a slope of $-\frac{1}{4}$. Just remember that:

$m_{line} \cdot m_{normal} = -1$

5. oh ok I thought i was right on the tangent equation, but thanks for the normal info and hopefully the trig limit info will help me out. thanks.

6. Originally Posted by Chop Suey
Multiply by $\frac{x}{x}$, so that you would get:

$\frac{\sin{4x}}{x} \cdot \frac{x}{\sin{3x}}$

What's left is simple manipulation to get your limit into something like sin(u)/u
would this still work with if the question were to be:
sin2x
x

7. $\frac{{\sin (4x)}}
{{\sin (6x)}} = \left( {\frac{4}
{6}} \right)\left( {\frac{{\frac{{\sin (4x)}}
{{4x}}}}
{{\frac{{\sin (6x)}}
{{6x}}}}} \right)$