Find each of the following Limits - algebraically - if they exist?

Question

The limit as alpha approaches 0, Sin 4 alpha / Sin 6 alpha.

Answer 1

For simplicity, I'm gonna use x instead of alpha.

lim (x => 0, sin4x / sin6x)

First, read up on L'Hospital's rule, which states you can individually take the derivative of the numerator and denominator, and solve the new limit to get the answer. L'Hospital's rule only works if you obtain the form 0/0 or infinity/infinity after plugging in your limit.

So let's plug in the limit and see what we get.
If x = 0, then sin 4x / sin 6x = sin(0)/sin(0) = 0/0.

Because we have 0/0, we can use L'Hospital's rule.

lim (x => 0, sin4x / sin6x) = lim (x => 0, [cos4x * 4]/[cos6x * 6])

Note: When taking the derivative of those functions, you have to use the chain rule. That explains the 4 and the 6.

We want to pull the constants out of the limits, since we want them out of the way, and it is a perfectly legitimate thing to do when solving limits, derivatives, and integrals.

4/6 * lim (x => 0, cos4x / cos6x)

We plug in x = 0 to see what we get, and we get cos(0)/cos(0) = 1, so we can just plug in x normally and remove the limit.

4/6 * cos(4*0)/cos(6*0) = 4/6 * 1/1 = 4/6

Answer 2

All of those limits use "L'Hopital's Rule", which states that if the reduce of a(x)/b(x) as you frame of mind a volume is undefined (0/0 or Infinity/Infinity), then the reduce will be a'(x)/b'(x). So: a) lim x-->one hundred twenty 5; [dice-root(x) - 5] / (x-one hundred twenty 5) = 0/0; lim x-->one hundred twenty 5; [dice-root(x) - 5] / (x-one hundred twenty 5) = lim x-->one hundred twenty 5; d/dx([dice-root(x) - 5]) / d/dx (x-one hundred twenty 5) = lim x-%3

Answer 3

get your own home work. Did you want me to cheat for you.