Evaluate the limits: (1) limit as x approaches 3 of sqrt(x^2 + 6x - 2); (2) limit as x approaches 3 of (x^2 + 2x - 15)/(x - 3); (3) limit as x approaches 4 of (sqrt(x) - 2)/(x - 4).

Step-by-step solution

1. 1

   Always try direct substitution first. If substituting the target value produces a defined number and the function is continuous there, that number is the limit. Only when substitution gives an indeterminate form such as $\tfrac00$ does further work become necessary.

2. 2

   (1) Substitution succeeds. The radicand at $x=3$ is $9+18-2=25>0$, and $\sqrt{\ }$ is continuous where its argument is positive, so

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

3. 3

   (2) Substitution gives 0/0, so factor. At $x=3$ the numerator is $9+6-15=0$ and the denominator is $0$. By the factor theorem $(x-3)$ divides the numerator:

   $x^{2}+2x-15=(x+5)(x-3)$

   For $x\neq 3$ the common factor cancels, and cancelling is legitimate precisely because a limit never evaluates the function at $x=3$:

   $\lim_{x\to 3}\frac{(x+5)(x-3)}{x-3}=\lim_{x\to 3}(x+5)=8$

4. 4

   (3) Substitution gives 0/0 with a radical, so use the conjugate. Multiply numerator and denominator by $\sqrt x+2$:

   $\frac{\sqrt x-2}{x-4}\cdot\frac{\sqrt x+2}{\sqrt x+2}=\frac{x-4}{(x-4)\left(\sqrt x+2\right)}$

   using $\left(\sqrt x-2\right)\left(\sqrt x+2\right)=x-4$.

5. 5

   Cancel and substitute.

   $\lim_{x\to 4}\frac{1}{\sqrt x+2}=\frac{1}{2+2}=\frac14$

   $\boxed{5,\qquad 8,\qquad \tfrac14}$

6. 6

   Check numerically. At $x=3.001$: $\sqrt{x^2+6x-2}=5.0012$ and $\tfrac{x^2+2x-15}{x-3}=8.001$. At $x=4.001$: $\tfrac{\sqrt x-2}{x-4}=0.24998$. All three approach the stated values. As a further check on (3), the limit is by definition the derivative of $\sqrt x$ at $x=4$, which is $\tfrac{1}{2\sqrt4}=\tfrac14$ ✓.