This is a perfect time to use l'Hospital's rule:
If a fraction becomes 0/0 at the limit (like this one does),
then the limit of the fraction is equal to the limit of
(derivative of the numerator)/(derivative of the denominator) .
I'm not sure why l'Hospital's rule stuck with me all these years.
But when it's appropriate, like for this one, you can't beat it.
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Since both the numerator and the denominator approach zero, the conditions for de l'Hospital's rule are fulfilled. Take the derivate of both the numerator and the denominator, and take the limit of the new fraction: lim (t2-2)/(t-4) = lim 2t / 1 = 8 / 1 = 8.
lim (x3 + x2 + 3x + 3) / (x4 + x3 + 2x + 2)x > -1From the cave of the ancient stone tablets, we cleared away several feet of cobwebs and unearthed"l'Hospital's" rule: If substitution of the limit results in ( 0/0 ), then the limit is equal to the(limit of the derivative of the numerator) divided by (limit of the derivative of the denominator).(3x2 + 2x + 3) / (4x3 + 3x2 + 2) evaluated at (x = -1) is:(3 - 2 + 3) / (-4 + 3 + 2) = 4 / 1 = 1
You can use the L'hopital's rule to calculate the limit of e5x -1 divided by sin x as x approaches 0.
0.5
The limit is the Golden ratio which is 0.5[1 + sqrt(5)]