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=Σ ( a ) x^a
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
