Find $a$ and $b$ such that $$ \lim_{x \to 0}\frac{\sqrt{ax+b}-2}{x}=1.$$
I'm not sure how to solve for two variables given that I only have one equation.
$\endgroup$ 43 Answers
$\begingroup$because the denumerator is $0$, the ambiguity should be of the type $\frac{0}{0}$ (because on any other case if the numerator is not zero the limit will be $\pm\infty$) so
$$\sqrt{a\times 0+b}-2=0\Rightarrow b=4$$
now we can use hopital theorem:
$$\lim_{x \to 0}\frac{\sqrt{ax+4}-2}{x}=\lim_{x\to 0}\frac{\frac{a}{2\sqrt{ax+4}}}{1}=1\Rightarrow a=4$$
Notice that \begin{align} & \frac{\sqrt{a x + b} - 2}{x} \\ = & \frac{(\sqrt{ax + b} - 2)(\sqrt{ax + b} + 2)}{x} \\ = & \frac{ax + b - 4}{x(\sqrt{ax + b} + 2)} \\ \end{align} Try $b = 4$ and $a = 4$.
$\endgroup$ 3 $\begingroup$$lim_{x \rightarrow 0} (\sqrt{ax+b}-2)=0 \Rightarrow b=4$ so we have
$lim_{x \rightarrow 0} \dfrac {(\sqrt{ax+4}-2)} x=lim_{x \rightarrow 0} \dfrac {(ax+4-4)} {x(\sqrt{ax+4}+2)}=lim_{x \rightarrow 0} \dfrac {a} {\sqrt{ax+4}+2}=\dfrac a 4 $
so we have $a=4$
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