Question

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


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Answer 1

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`