g(x)=[(ax+b)^(1/2)-1]/x for x not equal to 0 and g(x)=1 for x=0 if g(x) is continuous at x=0 then a+b=?

Question

amistre64 · Answer

Lhop perhaps?

anonymous · Answer

but the answer is given 3 i got 2

amistre64 · Answer

$$\lim\frac{(ax+b)^{1/2}-1}{x}$$
$$\lim\frac{a}{2(ax+b)^{1/2}}$$
$$\lim\frac{a}{2b^{1/2}}=1$$
a = 2sqrt(b); or b = a^2/4

did i mess that up any?

anonymous · Answer

well what after that

amistre64 · Answer

$$\frac{(2b^{1/2}x+b)^{1/2}-1}{x}$$
hmm, or maybe
$$\frac{(ax+b)^{1/2}-1}{x}~if~(ax+b)^{1/2}=x+1$$

amistre64 · Answer

$$ax+b = x^2+2x+1$$
$$b = x^2+(2-a)x+1$$
$$b+a = x^2+(2-a)x+1+a$$
at x=0, b+a = 1+a;  b=1
if b=1, then from the first run, b = a^2/4, if |a| = 2

amistre64 · Answer

so we could test out:  b = 1,  a = -2, or 2

anonymous · Answer

thanks a lot

amistre64 · Answer

youre welcome ... how did you try it?

anonymous · Answer

well i was stuck after applying l'hos

amistre64 · Answer

so was i :)  i spose since a=2sqrt(b), and b=1 .. a=2

i just had a hunch to try to make the top equal to x is all