Lim x-->0 ((sqrt (ax+b) - 2) / x) = 1 show steps ...

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