Respuesta :

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

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