find the limit of lim sin(x-1)/(x^2 + x - 2) x->1

without using l'hopitals rule

is the function, like this:

[sin(x-1)]/(x^2 + x - 2)
or this:
sin [(x-1)/(x^2 + x - 2)]
?

the first option

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. :)

To find the limit of the given expression without using L'Hôpital's rule, we can simplify the expression and then substitute the limit directly into the simplified equation. Here's how:

Step 1: Simplify the expression
We can factor the denominator, (x^2 + x - 2), to obtain (x - 1)(x + 2). Now, the expression becomes sin(x - 1) / ((x - 1)(x + 2)).

Step 2: Cancel out common factors
Since (x - 1) appears in both the numerator and denominator, we can cancel it out: sin(x - 1) / (x + 2).

Step 3: Substitute the limit
Now, we substitute the limit x -> 1 into the simplified expression:

lim (x -> 1) sin(x - 1) / (x + 2)

When x approaches 1, we have sin(0) / 3, which is simply 0 / 3 = 0.

Therefore, the limit of sin(x-1)/(x^2 + x - 2) as x approaches 1 is 0.