Finding the limit: $\lim_{x\to3^+} \frac{\sqrt{x-3}}{|x-3|}$

Question

Can anyone tell me how to properly solve this limit?

$\displaystyle \lim_{x\to3^+} \frac{\sqrt{x-3}}{|x-3|}$

I know the answer is positive infinity, and I would know how to do the problem if $x$ was just approaching $3$. Not sure how to go about doing this one though, since it's only from the positive side. My first thought was that I should ignore $-(x-3)$ when $x$ is less than $3$, but I'm not sure if I'm right. Any help please?

Answer 1

The function $x\mapsto\frac{\sqrt{x-3}}{|x-3|}$ is defined for $x>3$, and for every $x>3$ we have $$ \frac{\sqrt{x-3}}{|x-3|}=\frac{(\sqrt{x-3})^2}{(x-3)\sqrt{x-3}}=\frac{x-3}{(x-3)\sqrt{x-3}}=\frac{1}{\sqrt{x-3}}. $$ It follows that $$ \lim_{x\to3^+}\frac{\sqrt{x-3}}{|x-3|}=\lim_{x\to3^+}\frac{1}{\sqrt{x-3}}=\infty. $$

Answer 2

Condition $x\to3{+}\;$ means that $x>3$ and $x\ne{3}.$ Then $$\lim_{x\to3+} \frac{\sqrt{x-3}}{|x-3|}=\lim_{x\to3+} \frac{\sqrt{x-3}}{(\sqrt{x-3})^2}=\lim_{x\to3+} \frac{1}{\sqrt{x-3}}$$

Answer 3

you can think of it as $$\frac{(x-3)^{1/2}}{|x-3|}$$if you subtract the exponents by division rule, you get: $\frac{1}{2}-1 = -\frac{1}{2}$. The absolute symbol goes away because if the numerator is postive the whole function is postive because positive/positive is positive. Similarly, if the numerator is negative, its negative over positive so the answer is negative. After applying the difference exponent rule you get $$(x-3)^{-\frac{1}{2}}$$ and as $x \to 3^+$ it goes to infinity.