Expert: Paul Klarreich - 3/4/2010

Question
Without using L'Hopital rule limit x ->0  (tan(x) - sin(x))/x^3)
well I already tried it with the rule and got 1/2  but I am trying to do it without it as the assignment says
limit x ->0 [sin(x)/cos(x) - sin(x)] *(x^-3)

limit x ->0 [sin(x)/cos(x) - cos(x)*sin(x)/cos(x)] *(x^-3)

limit x ->0 [(sin(x) -  cos(x)sin(x))/cos(x)] *(x^-3)
limit x ->0 (sin(x)(1 -  cos(x)))/cos(x)] *(x^-3)



limit x ->0 (sin(x)(sin(x)^2 +cos(x)^2 -  cos(x)))/cos(x)] *(x^-3)


limit x ->0 (sin(x)(sin(x)^2 +cos(x)^2 -  cos(x)))/cos(x)] *(x^-3)
1-cos(x)=sin(x)^2 +cos(x)^2-cos(x)
1-cos(x)/cos(x) =tan(x)sin(x) + cos(x) -1
limit x ->0 (sin(x)(tan(x)sin(x) + cos(x) -1))/cos(x)] *(x^-3)
limit x ->0 (tan(x)(tan(x)sin(x) + cos(x) -1))/] *(x^-3)
limit x ->0 ((tan(x)^2sin(x) + cos(x)tan(x) -tan(x)))/] *(x^-3)
limit x ->0 ((sin(x)^3/cos(x) + sin(x) -tan(x)))/] *(x^-3)
limit x ->0 ((sin(x)^3/cos(x)  *(x^-3) + sin(x)*(x^-3) -tan(x)))*(x^-3)]
limi x->0 sin(x)^3/cos(x)x^3 + well limit x->0 cos(x)=1 and I thought that since limit x->0 sin(x)/x =1 then its also true that limit x ->0 sin(x)^3/x^3=1 because we can make it limit x ->0 (sin(x)/x)^3 =>(1)^3 /1 = 1
I cant do the rest of the the 2nd and the third also I know it isnt right to distribute if its undetermined but that was all I can think of trigonometry gets hard without l'hopital. I know that this all could be wrong but what I wrote is about everything I know so I am just showing you that I tried everything I know so tell me what I do not know :)

Answer
Questioner: hamad
Country: Kuwait
Category: Advanced Math
Private: No
Subject: limit
Question: Without using L'Hopital rule limit x ->0  (tan(x) - sin(x))/x^3)
well I already tried it with the rule and got 1/2  but I am trying to do it without it as the assignment says


limit x ->0 [sin(x)/cos(x) - sin(x)] *(x^-3)

limit x ->0 [sin(x)/cos(x) - cos(x)*sin(x)/cos(x)] *(x^-3)

limit x ->0 [(sin(x) -  cos(x)sin(x))/cos(x)] *(x^-3)
limit x ->0 (sin(x)(1 -  cos(x)))/cos(x)] *(x^-3)



limit x ->0 (sin(x)(sin(x)^2 +cos(x)^2 -  cos(x)))/cos(x)] *(x^-3)


limit x ->0 (sin(x)(sin(x)^2 +cos(x)^2 -  cos(x)))/cos(x)] *(x^-3)
1-cos(x)=sin(x)^2 +cos(x)^2-cos(x)
1-cos(x)/cos(x) =tan(x)sin(x) + cos(x) -1
limit x ->0 (sin(x)(tan(x)sin(x) + cos(x) -1))/cos(x)] *(x^-3)
limit x ->0 (tan(x)(tan(x)sin(x) + cos(x) -1))/] *(x^-3)
limit x ->0 ((tan(x)^2sin(x) + cos(x)tan(x) -tan(x)))/] *(x^-3)
limit x ->0 ((sin(x)^3/cos(x) + sin(x) -tan(x)))/] *(x^-3)
limit x ->0 ((sin(x)^3/cos(x)  *(x^-3) + sin(x)*(x^-3) -tan(x)))*(x^-3)]
limi x->0 sin(x)^3/cos(x)x^3 + well limit x->0 cos(x)=1 and I thought that since limit x->0 sin(x)/x =1 then its also true that limit x ->0 sin(x)^3/x^3=1 because we can make it limit x ->0 (sin(x)/x)^3 =>(1)^3 /1 = 1


I cant do the rest of the the 2nd and the third also I know it isnt right to distribute if its undetermined but that was all I can think of trigonometry gets hard without l'hopital. I know that this all could be wrong but what I wrote is about everything I know so I am just showing you that I tried everything I know so tell me what I do not know :)
............................................

OK, you have:

limit x ->0 [sin(x)/cos(x) - sin(x)]
           -------------------------
                  x^3


Henceforth,  limit x ->0  is written LIM (saves typing)

LIM  [sin(x) - sin(x) cos x]
    -------------------------
            x^3 cos x

LIM sin x   [1 - cos x]
   ------ -------------
    x       x^2 cos x


1 * LIM  [1 - cos x]
        ------------
          x^2 cos x


LIM  [1 - cos x] [1 + cos x]
    ----------- -----------
     x^2 cos x  [1 + cos x]

LIM        sin^2 x
    ----------------------
     x^2 cos x [1 + cos x]


LIM  sin x  sin x     1
    ------------- -------------------
     x       x     cos x [1 + cos x]

       LIM     1
1 * 1 *      -------------------
            cos x [1 + cos x]


LIM      1
     -------------------
     cos x [1 + cos x]

AND NOW (need some dramatic music here)
      1
------------------
cos 0 (1 + cos 0)


   1
---------- = 1/2
1 (1 + 1)

Cute.