Lim x→0 (√(ax+b)-2)/x=1
You want to know the value of "a" and "b"
lim x→0 (√(ax+b)-2)/x=(√(0+b)-2)/0=(√b -2)/0;
Then if (√b -2)/0=1; the numerator must be "0"
(√b-2)=0
√b=2
(√b)²=2²
b=4
It is necessary the numerator must be "0", if the denominator is "0" and the result is equal a number.
Therefore:
lim (√(ax+4)-2)/x=1
x⇒0
I imagine you know Taylor Series.
√(ax+4)=(4(1+ax/4))¹/²=2(1+ax/4)¹/²
Remember:
(1/2)
(1+x)ᵃ=<span>Σ ( a ) x^a
</span>
In our case:
(1/2) (1/2) (1/2)
(1+ax/4)¹/²=( 0) (ax/4)⁰+( 1 ) (ax/4)¹+( 2) (ax/4)²+...
=1 +(1/2) ax/4 + -1/8 (ax/4)²+...
=1+ax/8-a²x²/128+...
Therefore:
lim (√(ax+4)-2)/x=lim [2(1+ax/8-a²x²/128+...)-2]/x=
x⇒0 x⇒0
lim [(2+ax/4-a²x²/64+...)-2]/x=
x⇒0
lim (ax/4-a²x²/64+...)/x=
x⇒0
lim x(a/4-a²x/64+...)/x=
x⇒0
lim (a/4-a²x/64+...)=(a/4-0-0-0-...)=4/a
x⇒0
Because:
lim (√(ax+4)-2)/x=1
x⇒0
Then:
4/a=1 ⇒ a=4
Answer: a=4; b=4
Answer:
<h2>
-8 +6=-2 </h2>
it took me a while, but yeah....
hope im right, and i really hope this helps.
Step-by-step explanation:
10 because 10x1 is 10 and 5x2 is 10 which is the lowest common denominator for both.
Step-by-step explanation:
Given Area of Circle B = Area of Circle A
Given r = 1/2d = 1/2(16) = 8 in
and Area of Circle =
![\pi {r}^{2} \\ = (3.14) ({8}^{2} ) \\ = (3.14)(64) \\ = 200.96 {in}^{2}](https://tex.z-dn.net/?f=%5Cpi%20%7Br%7D%5E%7B2%7D%20%5C%5C%20%20%3D%20%283.14%29%20%28%7B8%7D%5E%7B2%7D%20%29%20%5C%5C%20%20%20%3D%20%283.14%29%2864%29%20%5C%5C%20%20%3D%20200.96%20%7Bin%7D%5E%7B2%7D%20)
I honestly don’t remember what a dilation is, but I can tell you that those triangles are not similar. 3:18 and 8:24 do not share the same ratio. 3 fits into 18 *6* times and 8 only fits into 24 *3*times.