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.
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if the numerator converges to something non-zero, what would be the limit of the fractions? – user251257 Aug 14 '15 at 18:27
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Have you considered trying some substitutions like $a=(x+4)$ and $b=4$ to see what could happen? – JB King Aug 14 '15 at 18:28
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How did you come up with those substitutions? – Jonathan Aug 14 '15 at 18:29
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you can use Hopital if only and if $\sqrt {ax+b}-2 =0$. Be careful. – 6-0 Aug 14 '15 at 18:29
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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:
$$\lim_{x \to 0}\frac{\sqrt{ax+4}-2}{x}=\lim_{x\to 0}\frac{\frac{a}{2\sqrt{ax+4}}}{1}=1\Rightarrow a=4$$
Sepideh Abadpour
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1Thank you. I have never actually learned hopital theorem. I should probably learn it. – Jonathan Aug 14 '15 at 18:31
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@Jonathan you say that the limit is equal to one. true? let us think numerator$\neq 0$ then it should be positive or negative. if numerator is positive when it tend to zero we will have $\frac{positive}{0^+}=+\infty$ or $\frac{positive}{0^-}=-\infty$ if the numerator is negative the $\frac{negative}{0^+}=-\infty$ or $\frac{negative}{0^-}=+\infty$ so in either case we will have $limit =\pm\infty$ so the numerator can not be positive or negative meaning that it's not nonequal to zero so it should be equal to zero – Sepideh Abadpour Aug 14 '15 at 19:14
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Thank you for that explanation. Does that mean that the limit is always in indeterminate form if the denominator is equal to zero and the limit does not tend toward infinity or if the numerator is zero and the limit doesn't tend toward zero? – Jonathan Aug 14 '15 at 19:56
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yes that's true. but they are necessary not sufficient conditions. – Sepideh Abadpour Aug 14 '15 at 20:05
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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 $a = 4$.
Zhanxiong
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From this form, I still don't understand how you see a and b equals 4. – Jonathan Aug 14 '15 at 18:33
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You must have $x$ cancelled so that $b = 4$. After that, you can see $a = 4$. – Zhanxiong Aug 14 '15 at 18:35
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$lim_{x \rightarrow 0} (\sqrt{ax+b}-2)=0 \Rightarrow b=4$ 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 $a=4$
Booldy
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