find a, b satisfy $\lim_{x\rightarrow 0}\dfrac{\sqrt{ax+b}-2}{x}=1$
Please, help

Question

find a, b satisfy $\lim_{x\rightarrow 0}\dfrac{\sqrt{ax+b}-2}{x}=1$
Please, help

loser66 · Answer

@agent0smith

agent0smith · Answer

The numerator and denominator must have the same value, for it to equal 1. So set them equal to each other, and plug in x=0.

loser66 · Answer

How can we say so? it is a limit, not a fraction.

loser66 · Answer

$f'(0) =\lim_{x\rightarrow 0}\dfrac{f(x+0) -f(0)}{x-0}=1$

So, $f(x) =\sqrt{ax+b}$, $f(0) =2, f'(0) =1$

$f'(x) =\dfrac{a}{2\sqrt{ax+b}}$

$f'(0) = \dfrac{a}{2b}=1$, so a =2b

$f(0) =2 =\sqrt b$, Hence, b=4, a=8

loser66 · Answer

However, when I plug back to get the lim, I got it wrong.

imqwerty · Answer

*$f'(0) = \dfrac{a}{2\color{red}{\sqrt{b}}}=1$

loser66 · Answer

Bingo!! thank you

loser66 · Answer

Now, everything go right!!

imqwerty · Answer

=]

agent0smith · Answer

What I said works for getting b. If the value of the numerator is not 0, then the limit would be infinity.

agent0smith · Answer

Since $$\large \dfrac{\sqrt{ax+b}-2}{x}$$ must be either of the form $\infty/\infty$ or 0/0 for it to be equal to 1. And we know it's not $\infty/\infty$...

loser66 · Answer

Got you. Thank you so much.