Question

$\displaystyle = \lim_{x\to3} \left[ \frac{\sqrt{2x+6}}{x} \right]$

Answer 1

$\frac{2\sqrt{3}}{3}$

Answer 2

We want to evaluate the limit:

$\displaystyle L = \lim_{x\to3} \left[ \frac{\sqrt{2x+6}}{x} \right]$

The function is $f(x) = \frac{\sqrt{2x+6}}{x}$.

We can try direct substitution of $x=3$ into the function.

Numerator: $\sqrt{2(3)+6} = \sqrt{6+6} = \sqrt{12}$

Denominator: $3$

Since the denominator is non-zero at $x=3$ and the numerator is defined (the expression inside the square root is positive), the function is continuous at $x=3$.

Therefore, the limit can be found by direct substitution:

$\displaystyle L = \frac{\sqrt{2(3)+6}}{3} = \frac{\sqrt{6+6}}{3} = \frac{\sqrt{12}}{3}$

Now, we simplify the expression:

$\sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3}$

So, the limit is:

$\displaystyle L = \frac{2\sqrt{3}}{3}$

The limit is evaluated by direct substitution of $x=3$ into the expression, as the function is continuous at $x=3$. Substituting $x=3$ gives $\frac{\sqrt{2(3)+6}}{3} = \frac{\sqrt{12}}{3}$. Simplifying $\sqrt{12}$ to $2\sqrt{3}$ gives the final result $\frac{2\sqrt{3}}{3}$.