Find the limit of sqrt(x^2 + 6x - 2) as x approaches 3.

Step-by-step solution

1. 1

   Test the radicand at the target point. Evaluate the inside first, because the whole question is whether the square root is defined and continuous there:

   $x^2+6x-2 \Big|_{x=3} = 9 + 18 - 2 = 25$

   This is positive, so $\sqrt{\cdot}$ is continuous in a neighbourhood of the point and substitution is legitimate.

2. 2

   Move the limit inside the root. For a continuous outer function,

   $\lim_{x\to 3}\sqrt{g(x)} = \sqrt{\lim_{x\to 3} g(x)}$

   This is the composition law, and it is the step that makes the problem a one-liner.

3. 3

   Substitute and simplify.

   $\sqrt{\lim_{x\to 3}(x^2+6x-2)} = \sqrt{25} = 5$

   The principal square root is taken, so the answer is $+5$, not $\pm 5$.

4. 4

   Note when this shortcut would fail. If the radicand had approached a negative value, the two-sided limit would not exist over the reals; if it had approached $0$, one would need to check that the function is defined on both sides. Neither happens here — the radicand $x^2+6x-2$ is positive for all $x > -3 + \sqrt{11} \approx 0.317$, comfortably including $x = 3$.

5. 5

   Confirm numerically. At $x = 2.99$: $\sqrt{8.9401+17.94-2} = \sqrt{24.8801} = 4.98800$. At $x = 3.01$: $\sqrt{9.0601+18.06-2} = \sqrt{25.1201} = 5.01199$. The two one-sided values bracket $5$ and close in on it ✓, and a symbolic limit returns exactly $5$.