1. Factorize the numerator:
x¹⁰ - 1 = (x - 1) (x⁹ + x⁸ + x⁷ + … + x² + x + 1)
Then
[tex]\displaystyle \lim_{x\to1} \frac{x^{10}-1}{x-1} = \lim_{x\to1} (x^9+x^8+\cdots+x+1) = \boxed{10}[/tex]
2. Recall that
[tex]\displaystyle \lim_{x\to0} \frac{1-\cos(x)}x = 0[/tex]
Then
[tex]\displaystyle \lim_{x\to0} \frac{1-\cos(2x)}x = 2 \lim_{x\to0} \frac{1-\cos(2x)}{2x} = 2\times0 = \boxed{0}[/tex]
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
[tex]\displaystyle f'(x) = \lim_{h\to0} \frac{f(x+h) - f(x)}h[/tex]
[tex]\displaystyle f'(x) = \lim_{h\to0} \frac{\left((x+h)^2 + 2(x+h)\right) - \left(x^2+2x\right)}h[/tex]
[tex]\displaystyle f'(x) = \lim_{h\to0} \frac{x^2 + 2xh + h^2 + 2x + 2h - x^2 - 2x}h[/tex]
[tex]\displaystyle f'(x) = \lim_{h\to0} \frac{2xh + h^2 + 2h}h[/tex]
[tex]\displaystyle f'(x) = \lim_{h\to0} (2x + h + 2)[/tex]
[tex]\displaystyle f'(x) = 2x+2[/tex]
and
[tex]\displaystyle f''(x) = \lim_{h\to0} \frac{f'(x+h) - f'(x)}h[/tex]
[tex]\displaystyle f''(x) = \lim_{h\to0} \frac{(2(x+h)+2) - (2x+2)}h[/tex]
[tex]\displaystyle f''(x) = \lim_{h\to0} \frac{2x + 2h + 2 - 2x - 2}h[/tex]
[tex]\displaystyle f''(x) = \lim_{h\to0} \frac{2h}h[/tex]
[tex]\displaystyle f''(x) = \lim_{h\to0} 2[/tex]
[tex]\displaystyle f''(x) = 2[/tex]
