samuelryancampbell
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Calculus Limit Question--please walk me through it so I can understand it!
- Thread starter samuelryancampbell
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[MATH]f(x) = 3x^2.[/MATH]
Ignore the limit aspect initially.
[MATH]h \ne 0 \text { and } x = 6 \implies\dfrac{f(x + h) - f(x)}{h} = \dfrac{f(6 + h) - f(6)}{h} = \dfrac{3(6 + h)^2 - 3 * 6^2}{h} =[/MATH]
[MATH]\dfrac{3(36 + 12h + h^2) - 3 * 36}{h} = \dfrac{3(36 - 36) + 36h + 3h^2}{h} = \dfrac{h(36 + 3h)}{h} = 36 + 3h.[/MATH]
There is NOTHING hard or mysterious about it. It is very simple algebra. Check it with an example.
If h = - 1 and x = 6, then 3x^2 = 108, 3(x - h)^2 = 3 * 5^2 = 75. The difference - 33. Divide that by - 1, and you get - 33. That is the same number you get from 36 + 3(-1) = x^2 + 3h.
So we can say
[MATH]f(x) = 3x^2, \ x = 6,\ \text { and }h \ne 0 \implies \lim_{h \rightarrow 0} \dfrac{f(6 + h) - f(6)}{h} = \lim_{h \rightarrow 0} (36 + 3h) = WHAT?[/MATH]
Do the algebra first before worrying about limits. It will likely seem way more intuitive.
Ignore the limit aspect initially.
[MATH]h \ne 0 \text { and } x = 6 \implies\dfrac{f(x + h) - f(x)}{h} = \dfrac{f(6 + h) - f(6)}{h} = \dfrac{3(6 + h)^2 - 3 * 6^2}{h} =[/MATH]
[MATH]\dfrac{3(36 + 12h + h^2) - 3 * 36}{h} = \dfrac{3(36 - 36) + 36h + 3h^2}{h} = \dfrac{h(36 + 3h)}{h} = 36 + 3h.[/MATH]
There is NOTHING hard or mysterious about it. It is very simple algebra. Check it with an example.
If h = - 1 and x = 6, then 3x^2 = 108, 3(x - h)^2 = 3 * 5^2 = 75. The difference - 33. Divide that by - 1, and you get - 33. That is the same number you get from 36 + 3(-1) = x^2 + 3h.
So we can say
[MATH]f(x) = 3x^2, \ x = 6,\ \text { and }h \ne 0 \implies \lim_{h \rightarrow 0} \dfrac{f(6 + h) - f(6)}{h} = \lim_{h \rightarrow 0} (36 + 3h) = WHAT?[/MATH]
Do the algebra first before worrying about limits. It will likely seem way more intuitive.