Calculus

find the limit. use L'Hopital's Rule if necessary.

lim (x^2+3x+2)/(x^2+1)
x -> -1

  1. 👍
  2. 👎
  3. 👁
  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

    1. 👍
    2. 👎
  2. So you just plug -1 into the equation to get the limit?

    1. 👍
    2. 👎
  3. isn't there a way you do it with the derivatives?

    1. 👍
    2. 👎
  4. 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

    1. 👍
    2. 👎

Respond to this Question

First Name

Your Response

Similar Questions

  1. Calculus

    lim (arccsinx)(cotx) x-->0+ How do I solve this limit using L'Hopital's rule?

  2. Calculus Limits

    Question: If lim(f(x)/x)=-5 as x approaches 0, then lim(x^2(f(-1/x^2))) as x approaches infinity is equal to (a) 5 (b) -5 (c) -infinity (d) 1/5 (e) none of these The answer key says (a) 5. So this is what I know: Since

  3. calculus again

    Suppose lim x->0 {g(x)-g(0)} / x = 1. It follows necesarily that a. g is not defined at x=0 b. the limit of g(x) as x approaches equals 1 c.g is not continuous at x=0 d.g'(0) = 1 The answer is d, can someone please explain how?

  4. calculus

    Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it. lim x→∞ {(5x − 4)/(5x + 3)}^5x + 1

  1. calculus

    Using L'Hôpital's rule, evaluate lim of xe^(-x) as x approaches infinity

  2. Calc 1

    Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it. lim x→0 (sin^−1 x)/7x Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method,

  3. Calculus

    Find the limit as x approaches infinity of sin(2x)/x The answer is 0. How can I show that this is an indeterminate form so I can use L'Hopital's? And even if I use L'Hopital's, it comes out to be 2cos(2x), which doesn't help if I

  4. Calculus

    Which of the following functions grows at the same rate as 3^x as x goes to infinity? a)2^x b)sqrt(3^(x)+4) c)sqrt(6^x) d)sqrt(9^(x)+5^(x)) I tried this, but I'm not sure how to deal with the square roots when using l'hopital's

  1. Calculus

    Use L'Hoplital's rule to find the limit. Lim x->0 (3-3cos(x))/(sin(4x))

  2. precalcus

    can someone explain how to evaluate limits? i only learned how to solve the simpler ones that goes to infinity using the 1/x = 0 or 1/n = 0 rule. am i suppose to make the x whatever number they gave me? but then problem b would be

  3. calculus

    use L'Hopital's Rule to evaluate lim (4x(cos 8x-1))/(sin 8x - 8x) as x->0

  4. calc

    need to find: lim as x -> 0 of 4(e^2x - 1) / (e^x -1) Try splitting the limit for the numerator and denominator lim lim x->0 4(e^2x-1) (4)x->0 (e^2x-1) ______________ = ________________ lim lim x->0 e^X-1 x->0 e^x-1 Next solve for

You can view more similar questions or ask a new question.