Lim (x^2+2x+1)/(x^4-1) = 0/0 =====> using algebra
x->-1
Lim [ (x + 1) * (x+1) ] / [ (x-1) * (x+1) ]
Lim [ (x+1) ] / [ (x-1) ] = ( -1 + 1 ) / (-1-1) = 0 / -2 = 0
or using L'Hopial Rule
(2x + 2)/(4x^3)
Lim (2x + 2)/(4x^3) = ( 2*-1 + 2 ) / (4 *(-1)^3) = 0 / -4 = 0
Avoiding L'Hopitals rule:
lim x->-1 (x² + 2x + 1)/(x² + 1)(x² - 1)
lim x->-1 (x + 1)²/(x² + 1)(x + 1)(x - 1)
lim x->-1 (x + 1)/(x² + 1)(x - 1)
lim x->-1 1/(x² + 1) + 2/(x² + 1)(x - 1) = 1/2 - 1/2 = 0
l'Hopital's rule: if f(a)=g(a)=0, then lim xâa (f(x)/g(x)) = lim xâa (f'(x)/g'(x))
So:
lim xâ-1 ((x^2 + 2x + 1)/(x^4 - 1)) = lim xâ-1 ((2x + 2)/4x^3) = 0
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Verified answer
Lim (x^2+2x+1)/(x^4-1) = 0/0 =====> using algebra
x->-1
Lim [ (x + 1) * (x+1) ] / [ (x-1) * (x+1) ]
x->-1
Lim [ (x+1) ] / [ (x-1) ] = ( -1 + 1 ) / (-1-1) = 0 / -2 = 0
x->-1
or using L'Hopial Rule
(2x + 2)/(4x^3)
Lim (2x + 2)/(4x^3) = ( 2*-1 + 2 ) / (4 *(-1)^3) = 0 / -4 = 0
x->-1
Avoiding L'Hopitals rule:
lim x->-1 (x² + 2x + 1)/(x² + 1)(x² - 1)
lim x->-1 (x + 1)²/(x² + 1)(x + 1)(x - 1)
lim x->-1 (x + 1)/(x² + 1)(x - 1)
lim x->-1 1/(x² + 1) + 2/(x² + 1)(x - 1) = 1/2 - 1/2 = 0
l'Hopital's rule: if f(a)=g(a)=0, then lim xâa (f(x)/g(x)) = lim xâa (f'(x)/g'(x))
So:
lim xâ-1 ((x^2 + 2x + 1)/(x^4 - 1)) = lim xâ-1 ((2x + 2)/4x^3) = 0