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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
    therefore,
    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))
    1/3

    hope this helps. :)

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