trigonometric limits

• Oct 8th 2008, 05:31 AM
fortplainman
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
• Oct 8th 2008, 05:35 AM
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.
• Oct 8th 2008, 05:35 AM
Chop Suey
Multiply by $\displaystyle \frac{x}{x}$, so that you would get:

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

What's left is simple manipulation to get your limit into something like sin(u)/u
• Oct 8th 2008, 05:38 AM
Chop Suey
Quote:

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:

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

Where $\displaystyle (x_1, y_1)$ is the point in question and m is the slope. The equation of the normal line to this line contains $\displaystyle (x_1, y_1)$, 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 $\displaystyle -\frac{1}{4}$. Just remember that:

$\displaystyle m_{line} \cdot m_{normal} = -1$
• Oct 8th 2008, 05:40 AM
fortplainman
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.
• Oct 8th 2008, 05:46 AM
fortplainman
Quote:

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

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

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
• Oct 8th 2008, 06:46 AM
Plato
$\displaystyle \frac{{\sin (4x)}} {{\sin (6x)}} = \left( {\frac{4} {6}} \right)\left( {\frac{{\frac{{\sin (4x)}} {{4x}}}} {{\frac{{\sin (6x)}} {{6x}}}}} \right)$