lim x tends to 0 sin5x/tan3x
\[\large \lim_{x \rightarrow 0}\frac{\sin5x}{\tan3x} \qquad = \qquad \lim_{x \rightarrow 0}\frac{\sin5x}{\sin3x}\cos3x\]Understand what I did so far? I just rewrote tangent as sines and consines using the appropriate identity.
Here is an important identity that we can use,\[\large \lim_{\theta \rightarrow 0}\frac{\sin \theta}{\theta} \qquad = \qquad 1\]
the answer is 5/3 and i dnt know how do i get that
Yes I'm trying to explain that... Are you following along..?
yeah i am
Is it powers, or coefficients?
Umm I can't think of the right way to explain this. It's a bit of a tricky problem. We're going to split into a product of two limits. and we need to form a 5x on the bottom, below the sin5x. And a 3x above the sin3x so we can apply our identity. This might be a little confusing, see if you can follow these steps.
coefficients
nope i dint
Multiply the top and bottom by x, 5, and 3, \[\large \lim_{x \rightarrow 0}\frac{5\cdot3x\sin5x}{3\cdot5x\sin3x}\cos3x\] Which we can write like this, \[\large \dfrac{5}{3}\lim_{x \rightarrow 0}\frac{3x\sin5x}{5x\sin3x}\cos3x\]
Whoops...
\[\large \dfrac{5}{3}\lim_{x \rightarrow 0}\frac{\sin5x}{5x}\cos3x\cdot \lim_{x \rightarrow 0}\frac{3x}{\sin3x}\]
I did a bunch of shuffling things around. Like I said, this one is a bit tricky :\ you have to be comfortable with your multiplication.
what happens to cos 3x ?
Based on the identity posted earlier, we know that these orange parts will give us 1.\[\large \dfrac{5}{3}\color{orangered}{\lim_{x \rightarrow 0}\frac{\sin5x}{5x}}\cos3x\cdot \color{orangered}{\lim_{x \rightarrow 0}\frac{3x}{\sin3x}}\] And luckily the cosine is also approaching 1 as x approaches 0. So we end up with, \(\large \dfrac{5}{3}\cdot\color{orangered}{1}\)
You can't pull the cosine outside of the limit, since it includes x. But as long as we take the limit of it, we're fine. Cosine 0 is 1 yes? :D
yea thanks a lot
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