Question

Help with limits please!!!!

Answer 1

\[\lim_{x \rightarrow 0} x \csc 25x\]

Answer 2

i know that csc is 1/sin

Answer 3

Ever seen L'hopitals rule? Or is this calc 1?

Answer 4

calc 1

Answer 5

Well, we can be a bit sneaky with this in order to get what we want. So you may have seen it, but there's an identity that says:
\[\lim_{x \rightarrow 0}\frac{ sinx }{ x }=1\]Now this is true as long as the x in the angle of sin and the x in the denominator match. Doesnt matter what it is, if its sin(3x)/3x, its 1, sin(pix)/pix, still 1. So we need to take advantage of that identity, its just finding a way to manipulate this function as to get that. Now there are a couple ways, but this is how I think of it.

SO as you said, csc is 1/sin, so we can write:
\[\frac{ x }{ \sin25x }\]
Now here's where the trick comes in that I would use.
\[x = 1/(1/x)\]Because if you divided it out, you would flip and multiply and get back to x. So if I rewrite x in the funky way I have, I would have:
\[\frac{ 1 }{\frac{ \sin25x }{ x } } \]
Well, thats the form we wanted, to have sinx in a numerator and x in the denominator. Problem is we need the angle of sin to match the denominator, meaning that bottom has to be 25x, not just x. In order to get that to happen, we multiply top and bottom by 1/25, which gives us:

\[\frac{ \frac{ 1 }{ 25 } }{ \frac{ \sin25x }{ 25x } }\]And because
\[\lim_{x \rightarrow 0}\frac{ \sin25x }{ 25x }=1\], we are basically left with (1/25) divided by 1, or simply 1/25.

Answer 6

i think i understand it know thank you very much :)

Answer 7

Np : )