Question

Need help.... Solve this question step by step please.... Number 11 and 12 thank you

Answer 1

Limits.

$let\ A=\frac{x^2-3x+2-(x-1)*\sqrt{x^2-4}}{x^2+3x+2-(x+1)*\sqrt{x^2-4}}*\sqrt{\frac{x+2}{x-2}}\\\\Substitute\ x=2+h,\ as\ x \to\ 2,\ \ h \to 0\\\\A=\frac{4+h^2+4h-3(2+h)+2-(2+h-1)*\sqrt{2^2+h^2+4h-4}}{4+h^2+4h+3(2+h)+2-(2+h+1)*\sqrt{2^2+h^2+4h-4}}*\sqrt{\frac{2+h+2}{2+h-2}}\\\\A=\frac{h^2-\sqrt{h(h+4)}}{h^2+12+7h-(3+h)\sqrt{h(h+4)}}*\sqrt{\frac{4+h}{h}}\\\\strike\ off\ \ \sqrt{h}\ from\ numerator\ and\ denominator.\\\\A=\frac{h\sqrt{h}-\sqrt{h+4}}{h^2+12+7h-(3+h)\sqrt{h(h+4)}}*\sqrt{4+h}\\\\Substitute\ h=0$

$A=\frac{0-\sqrt{4}}{12}*\sqrt{4}\\\\A=-1/3$
=============================

$A=\lim_{x \to 1} \frac{\sqrt{ax+b}-\sqrt{x+1}}{x^2-1} =4\sqrt{2}\\\\rationalize\ numerator\\\\A=\lim_{x \to 1} \frac{[\sqrt{ax+b}-\sqrt{x+1}][\sqrt{ax+b}+\sqrt{x+1}]}{[\sqrt{ax+b}+\sqrt{x+1}](x-1)(x+1)}\\\\=\lim_{x \to 1} \frac{(ax+b)-(x+1)}{[\sqrt{ax+b}+\sqrt{x+1}](x-1)(x+1)}\\\\=\lim_{x \to 1} \frac{(a-1)x+b-1}{[\sqrt{ax+b}+\sqrt{x+1}](x-1)(x+1)}$

To have a limit for A existing at x -> 1,  the numerator has to have a factor x - 1.

   so          a - 1 = 1    and    b - 1 = -1
       so      a = 2    and  b = 0.

substitute in A and cancel (x - 1).    then substitute x = 1.

$A= \frac{1}{(\sqrt2+\sqrt2)2} = \frac{1}{4\sqrt2}$

that is the answer.  perhaps the given answer  in the question is not right.