一條p.math limit

lim[x→∞] [(x² - 2x - 3) / (x² -3x - 28) ]^ x
let y=[(x² - 2x - 3) / (x² -3x - 28) ]^ x
ln y
= x ln [(x² - 2x - 3) / (x² -3x - 28) ]
=x ln [1+(x+25)/ (x² -3x - 28) ]
=ln [1+(x+25)/ (x² -3x - 28) ] / (1/x)
lim[x→∞] ln y
= lim[x→∞] ln [1+(x+25)/ (x² -3x - 28) ] / (1/x)
=lim[x→∞] [(x^2-3x-28)/ (x² -2x - 3) ] [(-x^2-50x+47)/ (x² -3x - 28)^2 ] / (-1/x^2) [L'Hospital's rule]
=lim[x→∞] [ x^2 (x² +50x - 47) ] / [ (x² -3x - 28)(x^2-2x-3) ]
=1 (compare the coefficient of x^4)
So
lim[x→∞] ln y = 1
lim[x→∞] y =e^1 = e
lim[x→∞] [(x² - 2x - 3) / (x² -3x - 28) ]^ x = e