Question

1. -/1 points this is similar to section 2.2 problem 34: use algebraic simplifications to find the limit. \\( \lim_{h \to 0} \frac{2(x + 2h)^2 - 2x^2}{h} \\) answer: \\( \square \\) hint: follow example 11.

Explanation:

Step1: Expand the numerator

First, we expand \((x + 2h)^2\) using the formula \((a + b)^2 = a^2 + 2ab + b^2\). So, \((x + 2h)^2 = x^2 + 4xh + 4h^2\). Then multiply by 2: \(2(x + 2h)^2 = 2(x^2 + 4xh + 4h^2) = 2x^2 + 8xh + 8h^2\). Now subtract \(2x^2\) from this: \(2(x + 2h)^2 - 2x^2 = 2x^2 + 8xh + 8h^2 - 2x^2 = 8xh + 8h^2\).
The expression becomes \(\lim_{h \to 0} \frac{8xh + 8h^2}{h}\)

Step2: Simplify the fraction

Factor out \(h\) from the numerator: \(8xh + 8h^2 = h(8x + 8h)\). Then divide by \(h\) (assuming \(h
eq 0\), which is valid when taking the limit as \(h \to 0\) since we are approaching 0, not evaluating at 0): \(\frac{h(8x + 8h)}{h} = 8x + 8h\) (for \(h
eq 0\)).

Step3: Evaluate the limit

Now we take the limit as \(h \to 0\) of \(8x + 8h\). As \(h\) approaches 0, \(8h\) approaches 0, so the limit is \(8x + 0 = 8x\).

Answer:

\(8x\)