Mathematics Asked on December 15, 2020

My answer

Let $sqrt{ax+b}=y$

Then

$$lim_{yto sqrt b} frac{(y-1)a}{y^2-b}$$

Let $b=1$

Then $$lim _{yto 1} frac{a}{frac{y^2-1}{y-1}}$$

$$=frac a2 =1$$

$$a=2$$

The answer is correct, but this relies on assuming $b=1$, and that doesn’t seem appropriate. What is the correct answer for this?

$$frac{sqrt{ax+b}-1}{x}=frac{ax+b-1}{x(sqrt{ax+b}+1)}.$$ Now, we see that we need $b=1$ (otherwise, the limit does not exist) and $frac{a}{sqrt{b}+1}=1,$

which gives also $a=2$.

If $bneq1$ for $xrightarrow0$ we obtain: $$frac{sqrt{ax+b}-1}{x}=frac{ax+b-1}{x(sqrt{ax+b}+1)}=frac{a}{sqrt{ax+b}+1}+frac{b-1}{x(sqrt{ax+b}+1)}.$$ We see that for $b>0$ $$frac{a}{sqrt{ax+b}+1}rightarrowfrac{a}{sqrt{b}+1},$$ but $$ lim_{xrightarrow0}frac{b-1}{x(sqrt{ax+b}+1)}$$ does not exist.

Correct answer by Michael Rozenberg on December 15, 2020

Since $$lim_{xto 0}(sqrt{ax+b}-1) = lim_{xto 0}left(xfrac{sqrt{ax+b}-1}{x}right) =0$$

we get

$$lim_{xto 0}sqrt{ax+b}= 1Rightarrow b=1$$

The limit itself is now the first derivative of $sqrt{ax+1}$ at $x=0$:

$$left(sqrt{ax+1}right)'(0) = left. frac a{2sqrt{ax+1}}right|_{x=0} = frac a{2}stackrel{!}{=}1 Rightarrow a=2$$

Answered by trancelocation on December 15, 2020

We can solve this by using series expansion for $sqrt{ax+b}$. By expanding it, we have $$ sqrt{ax +b}= sqrt{b} +frac{ax}{2sqrt{b}} - frac{a^2x^2}{8b^{3/2}}+ {large O} $$ (${large O}$ means other higher powers of $x$ terms).

$$ lim_{xto 0} frac{sqrt{ax+b} -1}{x}= frac{sqrt{b}-1}{x} + frac{a}{2sqrt{b}}$$ Now, you can see that for limit to exist we have to have $sqrt{b}=1 implies b=1$. And by doing that we find $$ 1= a/2 \ a=2$$

Hope it helps!

Answered by Knight wants Loong back on December 15, 2020

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