Ap Calculus limits?

Question

What is the Lim x--->0 if

[sinx/2x]+3x+4)

Answer 1

Have you learned of something called Lo'Hiptals rule (not sure about the spelling), if you have this problem because extremely easy.
lim x->0 sinx/2x + 3x + 4 (form 0/0) multiply both the 3x and 4 terms by 2x to get x->0 sinx/2x + 6x^2/2x + 8x/2x. Now you can combine all the terms over the same demoninator. lim x->0 (sinx + 6x^2 + 8x)/2x (form 0/0) now we apply L'H's and get lim x->0 (cosx + 12x + 8)/2. This limit can be evaluated (1 + 0 + 8)/2 = 9/2

If you have not yet learned L'H then this is more complicated problem entirely. You most likely need to apply a rule that wil change the sinx into a cosx.

Answer 2

Simplify sinx / 2x into (1/2)(sinx / x), and knowing that lim --> 0 sinx/x = 1:

lim x --> 0 [(sinx / 2x) + 3x + 4]
lim x --> 0 [(1/2)(sinx / x) + 3x + 4]
1/2 + 4
9/2

Answer 3

Obviously, 4 --> 4 and 3x --> 0.

To find the limit for sinx/2x, you can use L'Hopital's rule: differentiate top and bottom: to get 0.5.

So the total answer is 4.5

Answer 4

All of those are certainly a similar. in basic terms use algebra and factoring. e) actuality (sqrt(x) - 3) * (sqrt(x) + 3) = (sqrt(x))^2 - 3^2 = x - 9 cut back (9-x) / 2sqrt(x) -6) = (9-x) / (2)(sqrt(x) - 3) x--9 From the reality we now get (3-sqrt(x)) * (3 + sqrt(x)) / -[2 * (3 - sqrt(x))] After the cancellation we've [3 + sqrt(x)] / 2 cut back [3 + sqrt(x)] / 2 = (3 + 3) / 2 = 6/2 = 3 x-->9 each and each of something use algebra in basic terms like this. considering which you asked 4 questions in one, i will leave something to you. they're all certainly a similar.