lim x>0 sin5x/sin4x

[Oneiromancy]

What the title says, find the limit.

If the denominator was say, 4x, I would have no problem doing this but the sin x function messed up I don't know what to do.

 

[stapel]

If the denominator was say, 4x, I would have no problem doing this

Okay, what would you have done, where that the limit?

Please be complete. Thank you!

Eliz.

 

[jwpaine]

When the limit can't be evaluated by simply plugging in the value, you will have to do some manipulation to get it into determinate form.

Look into trig identities or use L'Hopitals rule.

Tired, grumpy, and off to bed..... best of luck!
John

 

[Oneiromancy]

I'm only in Cal I so L'Hopital's rule is off limits.

If the problem were sin 5x / 4x I would have multiplied by a factor of one by using (5/4)/(5/4) so that it become sin 5x / 5x so now the problem becomes

1.25 * limit x > 0 sin 5x / 5x = 5/4

 

[o_O]

You can apply that exact same thing to this problem as well. Try multiplying by:

\(\displaystyle \frac{5x \cdot 4x}{5x \cdot 4x}\)

Can you see what you can do with this using the identity:

\(\displaystyle \lim_{x \to 0} \frac{sinx}{x} = 1\)

\(\displaystyle \lim_{x \to 0} \frac{x}{sinx} = \lim_{x \to 0} \frac{1}{\frac{sinx}{x}} = \frac{1}{1} = 1\)

 

[Oneiromancy]

5x * 4x wouldn't get rid of the denominator?

 

[o_O]

\(\displaystyle \frac{sin(5x)}{sin(4x)} \cdot \frac{5x4x}{5x4x} \quad = \quad \frac{sin(5x)}{5x} \cdot \frac{4x}{sin(4x)} \cdot \frac{5x}{4x}\)

 

[Oneiromancy]

You gave it away. The answer is 5x / 4x. Right? 4x / sin 4x = 1 / (sin 4x / 4x)

Thanks.

 

[Deleted member 4993]

You gave it away. The answer is 5x / 4x. Right? ->> Yes and No!!

4x / sin 4x = 1 / (sin 4x / 4x)

Thanks.