Question

Evaluate the following limit
$\underset{x\to 3}{\mathop{\lim }}\,\dfrac{\sqrt{2x+3}}{x+3}$

Answer 1

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

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