calculus... i really appreciate the help (limits)?

Question

lim as x goes to 4 of (4 - x) / (square root of x - 2)

lim as x goes to zero of (x - tan x) / (x - sin x)

lim as x goes to zero of (1 - cos^2 (2x) ) / x^2

if you could explain any of these it would be so great!!! thank you so much in advance.

for the first problem its just the square root of x. not the square root of (x - 2)

havent learned l'hospital's rule... could some one walk me through how to solve the last two?

Answer 1

(4-x)/sqrt(x)-2
multiply by sqrt(x) +2 /sqr(x) + 2
(4-x)(sqrt(x))/x-4
factor out a -1
(-1)(x-4)(sqrt(x))/x-4
the x-4 cancels out to get
-1*sqrt(x)
This is -2 when x = 4

For the last one use sin^2(x) + cos^2(x) = 1
You can replace x with 2x in this and it will still be true
sin^2(2x) + cos^2(2x) = 1
1 - cos^2(2x) = sin^2(2x)
substitute this in your original problem to get
sin^2(2x) / x^2
multiply by 4/4
4*sin^2(2x)/ (4*x^2) = (2 sin(2x)/(2x))^2
you can substitute u = 2x in the problem
as x ---> 0 , u = 2x also approaches 0
now you have the lim as u ---> 0 of
(2 sin(u)/u)^2
lim sin(u)/u = 1
so your limit is 4

Answer 2

you should shouse l'hospital rule to solve all the problems, that is take the derivatives of top and bottom term with respect to x until you can identify the limit (i.e you should avoid 0/0 zero/zero term).

e.g lim as x goes to zero of (1 - cos^2 (2x) ) / x^2
derivative of top term 0-(2cos2x . -sin2x .2)
derivative of bottom term = 2x

if you substitute x with 0, you will get 0/0 term which we avoid.

take the derivatives again of top and bottom term

derivative of top term = 4 cos 2x sin 2x which is equal to 2 sin 4x
derivative of bottom term = 2

as a result, you get :

lim as x goes to zero of 2 sin 4x / 2x , divide the top and bottom term with 2, you'll get

lim as x goes to zero sin 4x / x,
you can solve this by 2 ways :

1. there is a rule that lim as x goes to zero sin x / x = x, so that the answer to above question is 4.

2. take the derivative of top and bottom once again as it still gives you 0/0 term when you substitute x with 0

derivative of top term = 4 cos 4x
derivative of bottom term = 1 so that

lim as x goes to zero (4 cos 4x) / 1, substitute x with 0 yields final result as 4.

Answer 3

In the first case, there's no problem since it becomes
0/√4 = 0/2 = 0

But in the next two cases, L'Hopitals rule will have to be used.


Doug

Answer 4

Use L'Hopital's rule and take the derivatives of top and bottom until the limit is identified.

The answer to the first question is -4, not 0.