Calculuse/ Trig

[Guest]

The directions are to "determine the limit of the trigonometric function (if is exists) - I am having difficulty coming up with the answer for these questions. Any help that you can provide will be appreciated! Thanks!


lim 1-tan x / sin x - cos x
x-> Pie / 4

and

lim sin 2x / sin 3x
x-> 0

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[stapel]

Note: "Pi" is a number; "pie" is something you eat.

lim 1-tan x / sin x - cos x
x-> Pie / 4

Does this mean:

. . . . .lim_[x->pi/4]1 - (tan(x)/sin(x)) - cos(x)

...or:

. . . . .lim_[x->pi/4]1 - tan(x)/[sin(x) - cos(x)]

...or:

. . . . .lim_[x->pi/4] [1 - tan(x)]/[sin(x) - cos(x)]

...or something else? How far have you gotten?

lim sin 2x / sin 3x
x-> 0

Rewrite as:

. . . . .sin(2x)/sin(3x) = [3x/2x][sin(2x)/sin(3x)]

. . . . .=[sin(2x)/2x][3x/sin(3x)]

Then use the limit they gave you for sin(@)/@ as @ goes to zero (and where "@" stands for "theta").

Eliz.

 

[Gene]

Re #2: Does the name L'Hopital mean any thing to you?

 

[soroban]

Hello, ajaramz!

. . . . . . . . . 1 - tan x
. . . lim . . --------------
. x->π/4 . sin x - cos x

. . . . . . . . . . . . . . 1 - (sin x)/(cos x)
The function is: . -----------------------
. . . . . . . . . . . . . . . . sin x - cos x

Multiply top and bottom by cos x:

. . . . . . cos x - sin x . . . . . . . . - (sin x - cos x) . . . . . . . -1
. . . ------------------------ . = . ------------------------- . = . -------
. . . cos x (sin x - cos x) . . . . cos x (sin x - cos x) . . . . .cos x

Now take the limit . . .