We are asked to find the limit of two expressions, if they exist. If the limit does not exist, we need to explain why. (a) Find $\lim_{h\to 0} \frac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h}$. (b) Find $\lim_{x\to 3} (2x + |x-3|)$.

1. Problem Description

We are asked to find the limit of two expressions, if they exist. If the limit does not exist, we need to explain why.

(a) Find $\lim_{h\to 0} \frac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h}$.

(b) Find $\lim_{x\to 3} (2x + |x-3|)$.

2. Solution Steps

(a) We have:

$\lim_{h\to 0} \frac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h} = \lim_{h\to 0} \frac{\frac{x^2 - (x+h)^2}{x^2(x+h)^2}}{h} = \lim_{h\to 0} \frac{x^2 - (x^2 + 2xh + h^2)}{h x^2 (x+h)^2} = \lim_{h\to 0} \frac{-2xh - h^2}{h x^2 (x+h)^2} = \lim_{h\to 0} \frac{h(-2x - h)}{h x^2 (x+h)^2} = \lim_{h\to 0} \frac{-2x - h}{x^2 (x+h)^2}$

Now, we can evaluate the limit by substituting $h=0$:

$\frac{-2x - 0}{x^2 (x+0)^2} = \frac{-2x}{x^2 x^2} = \frac{-2x}{x^4} = \frac{-2}{x^3}$.

So, $\lim_{h\to 0} \frac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h} = \frac{-2}{x^3}$.

(b) We need to evaluate $\lim_{x\to 3} (2x + |x-3|)$. Since we are taking the limit as $x$ approaches 3, we need to consider the behavior of $|x-3|$ as $x$ approaches 3 from the left and from the right.

If $x > 3$, then $|x-3| = x-3$, so

$\lim_{x\to 3^+} (2x + |x-3|) = \lim_{x\to 3^+} (2x + (x-3)) = \lim_{x\to 3^+} (3x - 3) = 3(3) - 3 = 9 - 3 = 6$.

If $x < 3$, then $|x-3| = -(x-3) = 3-x$, so

$\lim_{x\to 3^-} (2x + |x-3|) = \lim_{x\to 3^-} (2x + (3-x)) = \lim_{x\to 3^-} (x + 3) = 3 + 3 = 6$.

Since the left-hand limit and the right-hand limit are equal, the limit exists and is equal to

6. So, $\lim_{x\to 3} (2x + |x-3|) = 6$.

3. Final Answer

(a) The limit is $\frac{-2}{x^3}$.

(b) The limit is $6$.