Math CentralQuandaries & Queries


Question from Teri, a student:

Although I have this problem completely worked out in front of me I still cannot understand
how it was done. The problem is:
Find the limit.
lim x->0 sin2x/tan7x. I know I'm supposed to break up tan into sinx/cosx.
I did that. I end up with sin2xcos7x/sin7x. Then I'm supposed to break up that fraction and multiply the whole thing
by2/7. After I do that I'm left with too many 2/7 because the cos7x part doesn't come out. It's supposed to go to 1 but
I don't see how. Please help me as I have been on this problem for an hour now. Thank you so much!!!

Hi Teri,

The key here is

limit sin(q)/q

What this says is that if you have the sine of some expression divided by the same expression and you take the limit of this fraction as the expression goes to zero, the value is 1. In your problem you have both sin(2x) and sin(7x) and I would like to make them sin(2x)/2x and sin(7x)/7x so that I have the sine of an expression divided by the same expression. This is what I do

step 1

What I did was divide sin(2x) by 2x to get "the sine of some expression divided by the same expression", perform the same trick with sin(7x) and then to maintain the equality I needed to multiply by (2x)/(7x) = 2/7. Now when I take the limits I get


Thus the final value is

1 × 1 × 1 × 2/7 = 2/7.

I hope this helps,

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