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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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