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1.lim f(x)= (x¹⁰-1)/(x-1)

x--->1=....
2.lim f(x)= (1-cos2x)/x
x----0
3.f(x)=x²+2x---->f'(x)=.... ----->f''(x)=
please answer with the steps




User Allyraza
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1 Answer

5 votes

1. Factorize the numerator:

x¹⁰ - 1 = (x - 1) (x⁹ + x⁸ + x⁷ + … + x² + x + 1)

Then


\displaystyle \lim_(x\to1) (x^(10)-1)/(x-1) = \lim_(x\to1) (x^9+x^8+\cdots+x+1) = \boxed{10}

2. Recall that


\displaystyle \lim_(x\to0) \frac{1-\cos(x)}x = 0

Then


\displaystyle \lim_(x\to0) \frac{1-\cos(2x)}x = 2 \lim_(x\to0) (1-\cos(2x))/(2x) = 2*0 = \boxed{0}

3. If f(x) = x² + 2x, then f'(x) = 2x + 2 and f''(x) = 2. This follows from the power rule for differentiation.

If you wish to show this using the definition, then


\displaystyle f'(x) = \lim_(h\to0) \frac{f(x+h) - f(x)}h


\displaystyle f'(x) = \lim_(h\to0) \frac{\left((x+h)^2 + 2(x+h)\right) - \left(x^2+2x\right)}h


\displaystyle f'(x) = \lim_(h\to0) \frac{x^2 + 2xh + h^2 + 2x + 2h - x^2 - 2x}h


\displaystyle f'(x) = \lim_(h\to0) \frac{2xh + h^2 + 2h}h


\displaystyle f'(x) = \lim_(h\to0) (2x + h + 2)


\displaystyle f'(x) = 2x+2

and


\displaystyle f''(x) = \lim_(h\to0) \frac{f'(x+h) - f'(x)}h


\displaystyle f''(x) = \lim_(h\to0) \frac{(2(x+h)+2) - (2x+2)}h


\displaystyle f''(x) = \lim_(h\to0) \frac{2x + 2h + 2 - 2x - 2}h


\displaystyle f''(x) = \lim_(h\to0) \frac{2h}h


\displaystyle f''(x) = \lim_(h\to0) 2


\displaystyle f''(x) = 2

User Wolfgang Bures
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