find the limit. use L'Hopital's Rule if necessary.
lim (x^2+3x+2)/(x^2+1)
x -> -1
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
So you just plug -1 into the equation to get the limit?
isn't there a way you do it with the derivatives?
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
To find the limit of a function as x approaches a certain value, you can substitute the value into the function and evaluate it. However, in this case, substituting -1 directly into the function results in an indeterminate form (0/0). To resolve this, we can apply L'Hopital's Rule.
L'Hopital's Rule states that if the limit of f(x)/g(x) as x approaches a certain value is of the form 0/0 or ∞/∞, and both f'(x) and g'(x) exist and g'(x) is not zero in the neighborhood of the value, then the limit of f(x)/g(x) is the same as the limit of f'(x)/g'(x).
Let's apply L'Hopital's Rule to find the limit:
1. Take the derivative of the numerator and the denominator separately.
f(x) = x^2 + 3x + 2
f'(x) = 2x + 3
g(x) = x^2 + 1
g'(x) = 2x
2. Evaluate the derivatives at x = -1.
f'(-1) = 2(-1) + 3 = 1
g'(-1) = 2(-1) = -2
3. Take the limit of the derivative quotient:
lim (x -> -1) f'(x)/g'(x) = lim (x -> -1) 1/-2 = -1/-2 = 1/2
Therefore, the limit of (x^2 + 3x + 2)/(x^2 + 1) as x approaches -1 is 1/2.