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I want to find the number $a$ and $b$ such that $\lim_{x\to 0} \frac{\sqrt{ax+b}-2}{x}=1$. First of all, I know that $b$ has to be 4, because the limit of the numerator has to be zero because the denominator is zero when we take its limit. My problem is with the number $a$. I need help. Thanks a lot! |
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Multiplying by the conjugate gives: $$ \frac{\sqrt{ax+b}-2}{x}=\frac{ax+b-4}{x(\sqrt{ax+b}+2)}. $$ As you already observed, you need $b=4$, for otherwise the limit is infinite. Then you can simplify by $x$. Now the limit is: $$ \frac{a}{\sqrt{4}+2}=\frac{a}{4}. $$ So you want $a=4$. |
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As $x\rightarrow 0$, the square root behaves as $$\sqrt{ax+b}=\sqrt{b}\sqrt{1+\frac{a}{b}x} \sim \sqrt{b}\left(1+ \frac{1}{2}\frac{a}{b}x+O(x^2)\right).$$ Subtracting the $2$ and dividing by $x$, we have $$ \frac{\sqrt{ax+b}-2}{x}\sim\frac{\sqrt{b}-2}{x}+\frac{a}{2\sqrt{b}}+O(x). $$ For the first (diverging) term to vanish, you need $b=4$; to make the second (constant) term equal to $1$, you also need $a=4$. |
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When $a=b=4$. We can calculate the binomial series of $\sqrt{ax+b}$ and we find that $$\sqrt{ax+b} \sim \sqrt{b} + \frac{a}{2\sqrt{b}}x + \cdots$$ Using this approximation, we see that $$\frac{\sqrt{ax+b}-2}{x} \sim \frac{\sqrt{b}-2}{x} + \frac{a}{2\sqrt{b}} + \cdots $$ where the tail "$+\cdots$" consists of terms divisible by $x$. Assuming that $\sqrt{b} \neq 2$, the limit is undefined as $x \to 0$. If $\sqrt{b} = 2$, i.e. $b=4$ then we have: $$\frac{\sqrt{ax+4}-2}{x} \sim \frac{a}{4}-\frac{a^2}{64}x + \cdots $$ where the tail "$+\cdots$" consists of terms divisible by $x^2$. In this case, the limit as $x \to 0$ is $\frac{1}{4}a$ and so we need $a=4$. It follows that $$\lim_{x \to 0}\left(\frac{\sqrt{4x+4}-2}{x}\right) = 1 \, . $$ |
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