posted by .

find the limit of lim sin(x-1)/(x^2 + x - 2) x->1
without using l'hopitals rule

  • maths -

    is the function, like this:
    [sin(x-1)]/(x^2 + x - 2)
    or this:
    sin [(x-1)/(x^2 + x - 2)]

  • maths -

    the first option

  • maths -

    alright then. if using L'hopital's Rule is not allowed, then recall the property:
    lim (sin x)/x = 1 as x->0
    lim [sin(x-1)]/(x^2 + x - 2)
    we can factor the denominator:
    (x^2 + x - 2) = (x-1)(x+2)
    we can then re-write the limit as:
    lim [sin(x-1)]/[(x-1)(x+2)]
    then we group terms, such as this:
    lim {[sin(x-1)]/(x-1)} * lim (1/(x+2))

    notice that the first looks like the property lim (sin x)/x = 1 as x->0 ,,
    but we do shifting in x in order to have x approach 1,,
    lim {[sin(x-1)]/(x-1)} as x->1 = 1
    therefore, its limit is 1,, thus,
    1 * lim (1/(x+2)) as x -> 1
    1 * (1/(1+2))

    hope this helps. :)

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Calculus

    Find the limit. Use L'Hopitals Rule if necessary. lim (x^2+3x+2)/(x^2+x) x -> 0
  2. calculus

    I am studying for a final exam and in our review packet we have this question, which I am having trouble with: lim (x-->0) ((e^x)+x)^(1/x) I tried this problem two ways. The first was with L'Hopitals rule... lim(x-->0) (1/x)((e^x)+x)^(1/x-1) …
  3. maths

    find the limit as x approaches 0 of the function: (e^x.sinx)/x without using hopitals rule
  4. Calculus

    Find the positive integers k for which lim ->0 sin(sin(x))/x^k exists, and then find the value the limit. (hint:consider first k=0, then k=1. Find the limit in these simple cases. Next take k=2 and finally consder k>2 and find …
  5. Maths

    Hi there, I'm currently stuck on a maths question. Find the limit as x approaches 0 for (1+sinx)^cotx I've put logs on both sides and am attempting to use hopitals rule but don't know where to proceed from here.
  6. L'Hopitals Rule

    5) Use the L’Hopital’s method to evaluate the following limits. In each case, indicate what type of limit it is ( 0/0, ∞/∞, or 0∙∞) lim x→2 sin(x^2−4)/(x−2) = lim x→+∞ ln(x−3)/(x−5) …
  7. calculus

    using the squeeze theorem, find the limit as x->0 of x*e^[8sin(1/x)] what i did was: -1<=sin(1/x)<=1 -8<=8*sin(1/x)<=8 e^(-8)<=e^[8*sin(1/x)]<=e^(8) x*e^(-8)<=x*e^[8*sin(1/x)]<=x*e^(8) lim x->0 [x*e^(-8)] …
  8. Calculus

    Can you please help me get the solution to this limit without using squeeze theorem and l'hopitals rule lim x to 0 of x^3 sin(1/x) lim x to 0 of x^2 sin^2(1/x)
  9. calculus

    find the limit without using L'Hopital's Rule Lim(X->-4) (16-x^2 / x+4)
  10. calculus

    Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it. lim x→0 (e^x − e^−x − 2x)/(x − sin(x))

More Similar Questions