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Calculus problem on finding limits?
How do you find the limit as x -> -1 for the equation ((x^2 + 8)^(1/2) - 3) / (x + 1)
Can someone explain without using derivatives?
2 Answers
- Anonymous8 years agoFavorite Answer
Use L'Hopital's rule. Take the derivative of the numerator and the denominator:
d/dx (x^2 + 8)^(1/2) - 3 = (1/2)*(x^2+8)^(-1/2)*2x
d/dx x + 1 = 1
So plug in x = -1:
(1/2)*((-1)^2+8)^(-1/2)*2*(-1)
=(1/2)*(9)^(-1/2)*-2
=-1*(1/3) = -1/3
The limit is -1/3
- 8 years ago
plug in x = -1 + h to the equation
why? because that's the limit of tangent line, lim as h->0 f(a+h)-f(a) / h
something like that (I forget, its been awhile)
so your equation should be lim h->0 ( your original equation with x = -1 +h)
and your answer should be -1/3
just like the person above.