How do I prove that $\lim_{x\to 9} \sqrt{x}=3$ using epsilon-delta proof

Question

How do I prove that $$\lim_{x\to 9} \sqrt{x}=3$$ using epsilon-delta proof

Answer 1

By definition $\lim_{x\to c}f(x)=L$ means that for every $\epsilon>0$ there is a $\delta>0$, such that $|x-c|<\delta\ \implies\ |f(x)-L|<\epsilon$ for all $x$.

Thus, let $\epsilon>0$. Then we are looking for a $\delta$ such that $$|x-9|<\delta\ \implies\ |\sqrt{x}-3|<\epsilon$$ Now note the following equalities: $$|\sqrt{x}-3|<\epsilon$$ $$-\epsilon<\sqrt{x}-3<\epsilon$$ $$3-\epsilon<\sqrt{x}<3+\epsilon$$ $$(3-\epsilon)^2<x<(3+\epsilon)^2$$ $$(3-\epsilon)^2-9<x-9<(3+\epsilon)^2-9$$ So given $\epsilon>0$, let $\delta=\max{|(3\pm\epsilon)^2-9|}$. Then $|x-9|<\delta$ and working our way through the above equalities from top to bottom, we see: $$|x-9|<\delta\ \implies\ |\sqrt{x}-3|<\epsilon$$

Answer 2

Start with scratch work: Assume that $$ |\sqrt{x}-3|<\varepsilon. $$ Now, do some work to make this look like $|x-9|$ by multiplying by $|\sqrt{x}+3|$ on both sides to get $$ |x-9|<|\sqrt{x}+3|\varepsilon. $$ Then, we'll choose $\delta<|\sqrt{x}+3|\varepsilon$ so that if $|x-9|<\delta$, then $$ |x-9|<\delta<|\sqrt{x}+3|\varepsilon. $$ We still have a pesky $x$ on the the right-hand-side of the inequality $\delta<|\sqrt{x}+3|\varepsilon$. Since this is scratch work, we can add extra conditions as we wish. In this case, let's say that $\delta<5$. In this case, since we will be assuming that $|x-9|<\delta$, we know that $4<x<14$. Therefore, $|\sqrt{x}+3|$ is bounded below by $5$ (when $x$ approaches $4$, the value of the square root approaches $2$). Combining these facts, we want: $$ \delta<5\quad\text{and}\quad\delta<5\varepsilon. $$

Now, the proof begins: Let $\varepsilon>0$ and let $\delta<\min\{5,5\varepsilon\}$ (we use a minimum so that both of the inequalities that we found in the scratch work will apply). Let $|x-9|<\delta$. Since $\delta<5$, we know that $4<x<14$. By the monotonicity of the square root function, we know that $2<\sqrt{x}<\sqrt{14}$ and that $5<\sqrt{x}+3<\sqrt{14}+3$. Observe that $\sqrt{x}+3$ is positive (and never zero). Therefore, $5<|\sqrt{x}+3|<\sqrt{14}+3$.

Next, we use the second inequality. Since $|x-9|<5\varepsilon$, we know that $|\sqrt{x}-3||\sqrt{x}+3|<5\varepsilon$. Since $|\sqrt{x}+3|$ is not zero (from above), we can divide by it to get $$ |\sqrt{x}-3|<\frac{5\varepsilon}{|\sqrt{x}+3|}. $$ We know that the denominator of the RHS is at least $5$ (and may be larger). Therefore, you are dividing by at least $5$ and so $$ |\sqrt{x}-3|<\frac{5\varepsilon}{|\sqrt{x}+3|}<\frac{5\varepsilon}{5}=\varepsilon. $$ This completes the proof.