Please help with limit problems!!!?

Question

Member since: June 26, 2007
Total points: 420 (Level 2)
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how do you evaluate the limit of (lim x->0) tan3x / x , and (lim x->0) 1-cos3x / x^2, and ( lim x-> 0) sinx - tanx / sin^3x.

Answer 1

Remember that sin u/u if u==>0has limit 1
so lim tan 3x/x = lim sin 3x/3x *3x/xcos 3x and sin3x/3x has lim=1 and so cos 3x so there remains only 3 which is the lim
lim (1-cos u)/u^2=1/2 so (1-cos 3x)/9x^2 *9x^2/x^2 has lim9/2
(sinx-tan x)/sin^x)= sinx(1-1/cosx)/sin^3x== (cosx-1)/x^2*x^2/sin^x
(cosx-1)x^2 has lim=-1/2 and (x/sinx)^2 has lim =1 so the lim =-1/2

Answer 2

There are some ways to approach limit problems. there are 3 ways of doing this;
1. Direct substitution
2. Graph the equation
3. Create a Table.

Because we get undefined results for all 3 if we put 0 for x, Direct substitution will not work. The best thing might be to graph all 3 equations and look at where the values are going as x->0.