Question

Evaluate the following limit
$\underset{x\to 3}{lim}{\displaystyle \frac{\sqrt{2x+3}}{x+3}}$

Answer 1

Hint: Find Left Hand Limit and Right Hand Limit. Use $\underset{x\to a^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(a+h)$ and $\underset{x\to a^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(a-h)$. Hence check whether Left Hand Limit equals Right Hand Limit or not.
Hence check whether the limit exists or not.

Complete step-by-step solution -

Limit of a function: Limit of a function f(x) is said to be equal to a if at point x=b, for all $\epsilon >0$, there exists a $\delta >0$ such that $|f\left(x\right)-a|<\epsilon$|f(x)-a|
Properties of the limit of a function:
[1] If $\underset{x\to a}{lim}f\left(x\right)$ exists and $\underset{x\to a}{lim}g\left(x\right)$ exists, then so do the limits $\underset{x\to a}{lim}f\left(x\right)+g\left(x\right),\underset{x\to a}{lim}f\left(x\right)g\left(x\right)$ and are equal to $\underset{x\to a}{lim}f\left(x\right)+\underset{x\to a}{lim}g\left(x\right)$ and $\left(\underset{x\to a}{lim}f\left(x\right)\right)\left(\underset{x\to a}{lim}g\left(x\right)\right)$ respectively. If $\underset{x\to a}{lim}g\left(x\right)\ne 0$, then $\underset{x\to a}{lim}{\displaystyle \frac{f\left(x\right)}{g\left(x\right)}}$ also exists and is equal to ${\displaystyle \frac{\underset{x\to a}{lim}f\left(x\right)}{\underset{x\to a}{lim}g\left(x\right)}}$.
[2] If the limit of a function exists, then it is unique.
Here $f\left(x\right)={\displaystyle \frac{\sqrt{2x+3}}{x+3}}$ and b = 3.
Claim: $\underset{x\to 3}{lim}f\left(x\right)={\displaystyle \frac{1}{2}}$
Proof:
Consider $g\left(x\right)=\sqrt{2x+3}$ and $h\left(x\right)=x+3$
Claim 1: $\underset{x\to 3}{lim}g\left(x\right)=3$
Observe that $|\sqrt{2x+3}-3|=\left|{\displaystyle \frac{2x+3-9}{\sqrt{2x+3}+3}}\right|=\left|{\displaystyle \frac{2x-6}{\sqrt{2x+3}+3}}\right|=2\left|{\displaystyle \frac{x-3}{\sqrt{2x+3}+3}}\right|$
Since $\sqrt{x}\ge 0$, we have
$|\sqrt{2x+3}-3|=2\left|{\displaystyle \frac{x-3}{\sqrt{2x+3}+3}}\right|\le 2\left|{\displaystyle \frac{x-3}{3}}\right|\le {\displaystyle \frac{2}{3}}|x-3|$
Now for $\epsilon >0$, choose $\delta ={\displaystyle \frac{3}{2}}\epsilon$, we have
Whenever $|x-3|<\delta ={\displaystyle \frac{3}{2}}\epsilon \Rightarrow {\displaystyle \frac{2}{3}}|x-3|<\epsilon$
Bit since $|\sqrt{2x+3}-3|\le {\displaystyle \frac{2}{3}}|x-3|$, we have
$|\sqrt{2x+3}-3|<\epsilon$
Hence whenever $|x-3|<\delta$, we have $|\sqrt{2x+3}-3|<\epsilon$
Hence we have $\underset{x\to 3}{lim}\sqrt{2x+3}=3$
Claim 2: $\underset{x\to 3}{lim}h\left(x\right)=6$
Observe that $|x+3-6|=|x-3|$
So if we choose $\delta =\epsilon$, we have $\mathrm{\forall }\epsilon >0,\mathrm{\exists }\delta =\epsilon >0$ such that $|x+6-3|<\epsilon$ whenever $|x-3|<\delta$
Hence $\underset{x\to 3}{lim}h\left(x\right)=6$
Hence $\underset{x\to 3}{lim}f\left(x\right)={\displaystyle \frac{\underset{x\to 3}{lim}g\left(x\right)}{\underset{x\to 3}{lim}h\left(x\right)}}={\displaystyle \frac{3}{6}}={\displaystyle \frac{1}{2}}$
Hence proved.

Note: Alternatively, we have
$\underset{x\to 3+}{lim}f\left(x\right)=\underset{h\to 0}{lim}{\displaystyle \frac{\sqrt{2(h+3)+3}}{h+3+3}}=\underset{h\to 0}{lim}{\displaystyle \frac{\sqrt{2h+9}}{h+6}}={\displaystyle \frac{\sqrt{2\left(0\right)+9}}{0+6}}={\displaystyle \frac{3}{6}}={\displaystyle \frac{1}{2}}$ and $\underset{x\to 3^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}{\displaystyle \frac{\sqrt{2(3-h)+3}}{3-h+3}}=\underset{h\to 0}{lim}{\displaystyle \frac{\sqrt{9-2h}}{6-h}}={\displaystyle \frac{\sqrt{9-0}}{6}}={\displaystyle \frac{1}{2}}$
Hence LHL = RHL $={\displaystyle \frac{1}{2}}$
Hence the limit exists and is equal to ${\displaystyle \frac{1}{2}}$.