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Find the limit. Use L'Hopitals Rule if necessary. lim (x^2+3x+2)/(x^2+x) x -> 0Asked by Joe
find the limit. use L'Hopital's Rule if necessary.
lim (x^2+3x+2)/(x^2+1)
x -> -1
lim (x^2+3x+2)/(x^2+1)
x -> -1
Answers
Answered by
drwls
In this case you don't need L'Hopital's Rule. The denominator is not zero at x = -1
The function equals 0/2 = 0 at x=-1
The function equals 0/2 = 0 at x=-1
Answered by
Joe
So you just plug -1 into the equation to get the limit?
Answered by
Joe
isn't there a way you do it with the derivatives?
Answered by
Reiny
Joe, you dont' always have to use Calculus to do Limit questions, just like drwls said.
The first thing I do is sub the approach value into your expression, there are three possilbilities:
1. your get a real number as an answer as above, 0/2 is real.
That is the answer to that limit, write it down and you are done.
2. you get a/0, where a is not equal to zero.
This is undefined, and there is no limit .
3. you get 0/0
This is the classic case and that is where the Calculus comes in.
You may try to factor it, if it is a simple algebraic expression, I can guarantee you it will factor.
If the expression is transcendental, that is it contains logs, trig or some other weird mathematical operation you might want to use L'Hopital's rule
The first thing I do is sub the approach value into your expression, there are three possilbilities:
1. your get a real number as an answer as above, 0/2 is real.
That is the answer to that limit, write it down and you are done.
2. you get a/0, where a is not equal to zero.
This is undefined, and there is no limit .
3. you get 0/0
This is the classic case and that is where the Calculus comes in.
You may try to factor it, if it is a simple algebraic expression, I can guarantee you it will factor.
If the expression is transcendental, that is it contains logs, trig or some other weird mathematical operation you might want to use L'Hopital's rule
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