answer:To evaluate the limit, consider the following steps: 1. Multiply and divide the expression by the conjugate: n-sqrt{n^2+(alpha+beta)n+alphabeta} = frac{n^2 - (n^2+(alpha+beta)n+alphabeta)}{n+sqrt{n^2+(alpha+beta)n+alphabeta}} 2. Simplify the numerator and factor out n from the denominator: frac{-(alpha+beta)n-alphabeta}{n+sqrt{n^2+(alpha+beta)n+alphabeta}} 3. Let h = frac{1}{n} so that ntoinfty corresponds to hto0^+. Then the expression becomes: lim_{{hto0^+}}frac{-h(alpha+beta)-alphabeta h^2}{1+sqrt{1+(alpha+beta)h+alphabeta h^2}} 4. To find the limit, rationalize the numerator: lim_{{hto0^+}}frac{-(alpha+beta)h^2-(alpha+beta)h-alphabeta h^2}{h(1+sqrt{1+(alpha+beta)h+alphabeta h^2})} 5. Simplify and apply L'Hopital's rule (as the limit is of the form frac{0}{0}): lim_{{hto0^+}}frac{-(alpha+beta)h-alphabeta}{1+sqrt{1+(alpha+beta)h+alphabeta h^2}} 6. Taking the limit, we get: frac{-(alpha+beta)}{1+1} = -frac{(alpha+beta)}{2} Hence, the limit is lim_{{ntoinfty}}left(n-sqrt{n+alpha}sqrt{n+beta}right) = -frac{(alpha+beta)}{2}

answer:In the long run, the elasticity of demand becomes more elastic for most products. As demand becomes more elastic, consumers become more price-sensitive and are less willing to bear the burden of a tax. This means that over time, producers will tend to bear a larger share of the tax burden for products with elastic demand. Conversely, for products with inelastic demand, consumers will continue to bear a larger share of the tax burden even in the long run.

answer:{eq}begin{align*} displaystyle f'(x) & = frac{ sec x tan x+sec x tan^2 x - sec^3 x}{(1 , + , tan x)^2} & = frac{sec x (tan x + tan^2 x - sec^2 x)}{(1 , + , tan x)^2} & = frac{sec x (1 - cos^2 x)}{(1 , + , tan x)^2} left[mathrm{Trigonometric Identity:} tan^2 x + 1 = sec^2 xright] & = frac{sec x sin^2 x}{(1 , + , tan x)^2} left[mathrm{Trigonometric Identity:} cos^2 x = 1 - sin^2 xright] end{align*} {/eq} Therefore, the simplified derivative is: {eq}displaystyle f'(x) = frac{sec x sin^2 x}{(1 , + , tan x)^2} {/eq}

answer:According to Kirchhoff's current law, the algebraic sum of currents at a node (branch point) is zero. That is, the total current flowing into the node equals the total current flowing out of it. Given: - The current carried by the first wire into the branch, {I_1} = 12;{rm{A}} - The current carried by the second wire away from the branch, {I_2} = 10;{rm{A}} We can express the current carried by the third wire, {I_3}, as: {eq}begin{align*} {I_1} + {I_3} &= {I_2} 12;{rm{A}} + {I_3} &= 10;{rm{A}} {I_3} &= 10;{rm{A}} - 12;{rm{A}} {I_3} &= -2;{rm{A}} end{align*}{/eq} Hence, the current carried by the third wire into the branch point is -2;{rm{A}}, indicating a current flowing out of the branch point.