Let a = lim_{x to 1} left( frac{x}{ln x} - fr | vedclass

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(D) First,we evaluate the limits:$a = \lim_{x 	o 1} \frac{x^2 - 1}{x \ln x} = \lim_{x 	o 1} \frac{(x-1)(x+1)}{x \ln x}$. Using $\lim_{t 	o 0} \frac

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Nilpotent

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(D) First,we evaluate the limits:$a = \lim_{x 	o 1} \frac{x^2 - 1}{x \ln x} = \lim_{x 	o 1} \frac{(x-1)(x+1)}{x \ln x}$. Using $\lim_{t 	o 0} \frac{\ln(1+t)}{t} = 1$,let $x = 1+t$,then $a = \lim_{t 	o 0} \frac{t(t+2)}{(1+t) \ln(1+t)} = \frac{2}{1 \cdot 1} = 2$.$b = \lim_{x 	o 0} \frac{x(x^2 - 16)}{x(4 + x)} = \frac{-16}{4} = -4$.$c = \lim_{x 	o 0} \frac{\ln(1 + \sin x)}{x} = \lim_{x 	o 0} \frac{\ln(1 + \sin x)}{\sin x} \cdot \frac{\sin x}{x} = 1 \cdot 1 = 1$.$d = \lim_{x 	o -1} \frac{(x + 1)^3}{3(\sin(x + 1) - (x + 1))}$. Let $u = x+1$,then $d = \lim_{u 	o 0} \frac{u^3}{3(\sin u - u)}$. Using Taylor series $\sin u \approx u - \frac{u^3}{6}$,we get $d = \lim_{u 	o 0} \frac{u^3}{3(u - u^3/6 - u)} = \lim_{u 	o 0} \frac{u^3}{3(-u^3/6)} = \frac{1}{-1/2} = -2$.Thus,the matrix is $A = \begin{bmatrix} 2 & -4 \ 1 & -2 \end{bmatrix}$.Calculating $A^2 = \begin{bmatrix} 2 & -4 \ 1 & -2 \end{bmatrix} \begin{bmatrix} 2 & -4 \ 1 & -2 \end{bmatrix} = \begin{bmatrix} 4-4 & -8+8 \ 2-2 & -4+4 \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}$.Since $A^2 = O$,the matrix is nilpotent.

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