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March 6, 2015

Posted by **khela** on Thursday, October 28, 2010 at 9:58pm.

without using l'hopitals rule

- maths -
**jai**, Thursday, October 28, 2010 at 10:01pmis the function, like this:

[sin(x-1)]/(x^2 + x - 2)

or this:

sin [(x-1)/(x^2 + x - 2)]

?

- maths -
**khela**, Thursday, October 28, 2010 at 10:10pmthe first option

- maths -
**jai**, Thursday, October 28, 2010 at 10:40pmalright 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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