m2 limits

1) [(x - 7)/(x - 5)]^x = [1 - 2/(x - 5)]^x. Let x - 5 = y, so x = y + 5
so it is changed to lim y tends to infinity of (1 - 2/y)^(y + 5)
= (1 - 2/y)^y (1 - 2/y)^5.
= e^(-2)(1^5) = e^(-2).
2)
sin (x + 1)/(x^2 - 1). Let x + 1 = y, so when x tends to - 1, y tends to 0.
so it changed to sin y/[(x+ 1)(x - 1)] = sin y/[y(y - 2)] = (sin y/y)[1/(y - 2)]
= (1)[1/(0 - 2)] = - 1/2.
3)
x ln ( 1 + 5/x + 6/x^2) = ln [1 + 5/x + 6/x^2]^x.
(1 + 5/x + 6/x^2) = (x^2 + 5x + 6)/x^2 = (x + 2)(x + 3)/x^2
= [(x + 2)/x][(x + 3)/x] = (1 + 2/x)(1 + 3/x)
so (1 + 5/x + 6/x2)^x = (1 + 2/x)^x(1 + 3/x)^x
so limit when x tends to infinity = ln (e^2)(e^3) = ln e^2 + ln e^3 = 2 + 3 = 5.
4)
(1 + 2/x - 15/x^2) = (x^2 + 2x - 15)/x^2 = (x +5)(x - 3)/x^2
so [(x + 4)(1 + 2/x - 15/x^2)]/x = (x + 4)(x + 5)(x - 3)/x^3
= [(x + 4)/x][(x + 5)/x][(x - 3)/x] = (1 + 4/x)(1 + 5/x)(1 - 3/x)
So similar to (3) above, limit = (e^4)(e^5)(e^ - 3) = e^6.

hope can help



圖片參考：http://latex.codecogs.com/gif.latex?\large%20\lim_{x\to\infty}(\frac{x-7}{x-5})^x\\%20=\lim_{x\to\infty}\frac{(1+\frac{-7}{x})^x}{(1+\frac{-5}{x})^x}\\%20=\frac{e^{-7}}{e^{-5}}\\%20=e^{-2}



圖片參考：http://latex.codecogs.com/gif.latex?\large%20\lim_{x\to%20-1}\frac{\sin{(x+1)}}{x^2-1}\\%20=\lim_{x\to%20-1}\frac{\sin{(x+1)}}{(x+1)(x-1)}\\%20=-\frac{1}{2}



圖片參考：http://latex.codecogs.com/gif.latex?\large%20\lim_{x\to\infty}x\ln{(1+\frac{5}{x}+\frac{6}{x^2})}\\%20=\lim_{x\to\infty}\ln{((1+\frac{3}{x})(1+\frac{2}{x}))^x}\\%20=\ln(e^3e^2)\\%20=5



圖片參考：http://latex.codecogs.com/gif.latex?\large%20\lim_{x\to\infty}(\frac{(x+4)(1+\frac{2}{x}-\frac{15}{x^2})}{x})^x\\%20=\lim_{x\to\infty}((1+\frac{4}{x})(1+\frac{5}{x})(1-\frac{3}{x}))^x\\%20=e^4e^5e^{-3}\\%20=e^6