<s>[INST] Prove $f(x)=\sqrt{2x-6}$ is continuous at $x=4$ by using precise definition. Please help me get the answer to this question.
Prove $f(x)=\sqrt{2x-6}$ is continuous at $x=4$ by using precise definition. ($\epsilon-\delta$ definition of limits.)
 [/INST] We have $|x-4|< \delta$.
$|f(x)-f(4)|=|\sqrt {2x-6}-\sqrt {2}|=|\sqrt{2}(\sqrt{x-3}-1)|$
Now, multiplying the numerator and denominator by $(\sqrt{x-3}+1)$,
$=\sqrt{2} |\frac{x-4}{\sqrt{x-3}+1}$|
Now, ${\sqrt{x-3}+1}$ is always greater than or equal to $1$. Thus, $\frac{x-4}{\sqrt{x-3}+1}\leqslant (x-4)$.
Therefore, $|f(x)-f(4)|\leqslant \sqrt{2}|(x-4)| < \sqrt{2} \delta=\epsilon$
Thus, $f(x)$ is continuous at $x=4$.
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