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Use the definition of a limit to prove the following: $$\lim_{x\to4}\frac{x-2}{\sqrt{x}+2}=\frac{1}{2}$$ I'm trying to prove that: $$\forall\varepsilon>0,\exists\delta>0; \\~\\ 0<|x-4|<\delta\Longrightarrow\left|\frac{x-2}{\sqrt{x}+2}-\frac{1}{2}\right|<\varepsilon$$ I don't know what to do with $\frac{x-2}{\sqrt{x}+2}-\frac{1}{2}$ I've tried using the triangle inequality, but I get wrong results.

I know for now: $\frac{1}{\sqrt{x}+2}<\frac{1}{\sqrt{3}+2}$

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Note that\begin{align}\frac{x-2}{\sqrt x+2}-\frac12&=\frac{2x-\sqrt x-6}{2\sqrt x+4}\\&=\frac{2(x-4)-\left(\sqrt x-2\right)}{2\sqrt x+4}\\&=\left(\sqrt x-2\right)\frac{2\sqrt x+3}{2\sqrt x+4}\\&=(x-4)\frac{2\sqrt x+3}{\left(\sqrt x+2\right)\left(2\sqrt x+4\right)}\\&\leqslant(x-4)\frac{2\sqrt x+3}8,\end{align}and therefore$$\left|\frac{x-2}{\sqrt x+2}-\frac12\right|\leqslant|x-4|\frac{2\sqrt x+3}8.$$So, if $|x-4|<1$, you have $x<5$, and therefore\begin{align}\frac{2\sqrt x+3}8&<\frac{2\sqrt5+3}8\\&<\frac{2\times3+3}8\\&=\frac98\\&<2.\end{align}Therefore, given $\varepsilon>0$, if you take $\delta=\min\left\{1,\frac\varepsilon2\right\}$, you have$$|x-4|<\delta\implies\left|\frac{x-2}{\sqrt x+2}-\frac12\right|<\varepsilon$$

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In this specific case since the form is not undetermined (i.e. not $\frac 00$), I think it is easier to show separately $|(x-2)-2|\to 0$ and $|(\sqrt{x}+2)-4|\to 0$, and then that the ratio $\to \frac 24$, because rationalizing the denominator will not bring much simplification actually.

For the square root use $|\sqrt{x}-2|=\frac{|x-4|}{|\sqrt{x}+2|}$ and bound the denominator (e.g. $|x-4|<3\implies x>1$, etc.)

You used $x>3$ instead, it is not wrong, but more complicated, I'd rather have denominator $<\frac 13$ than $<\frac 1{\sqrt{3}+2}$.

In the end use or show that $x_n\to a$ and $y_n\to b\neq 0$ then $\dfrac{x_n}{y_n}\to \dfrac{a}{b}$.

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