Question 254557
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{h\rightarrow 0}\,\,\frac{\left(x\,+\,h\right)^3\ -\ x^3}{h}]


Expand the binomial:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{h\rightarrow 0}\,\,\frac{x^3\,+\,3x^2h\,+\,3xh^2\,+\,h^3\ -\ x^3}{h}]


Collect terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{h\rightarrow 0}\,\,\frac{3x^2h\,+\,3xh^2\,+\,h^3}{h}]


Factor the numerator:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{h\rightarrow 0}\,\,\frac{h\left(3x^2\,+\,3xh\,+\,h^2\right)}{h}]


Eliminate the common factor in the numerator and denominator:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{h\rightarrow 0}\,\,3x^2\,+\,3xh\,+\,h^2]


Evaluate the limit:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{h\rightarrow 0}\,\,3x^2\,+\,3xh\,+\,h^2\ =\ 3x^2]


In general:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{h\rightarrow 0}\,\,\frac{\left(x\,+\,h\right)^n\ -\ x^n}{h}\ =\ nx^{n-1}]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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