Solution:
Given $\lim\limits_{\text{x} \rightarrow 0}\frac{(1+\text{x})^{\text{n}}-1}{\text{x}}=\lim\limits_{\text{x} \rightarrow 0}\frac{(1+\text{x})^{\text{n}}-(1)^{\text{n}}}{(1+\text{x})-(1)}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{(1+\text{x})^{\text{n}}-(1)^{\text{n}}}{1+\text{x}-(2)}=\text{n}(1)^{\text{n}-1}$
$=\text{n}$
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The ratio in which the line 3x + 4y + 2 = 0 divides the distance between the lines 3x + 4y + 5 = 0 and 3x + 4y - 5 = 0 is:
Let S = set of points inside the square, T = the set of points inside the triangle and C = the set of points inside the circle. If the triangle and circle intersect each other and are contained in a square. Then
$\text{S}\cap\text{T}\cap\text{C}=\phi$
$\text{S}\cup\text{T}\cup\text{C}=\text{C}$
$\text{S}\cup\text{T}\cup\text{C}=\text{S}$
$\text{S}\cup\text{T} = \text{S}\cap\text{C}$