Help with a calculus limit problem?

Question

lim (sin(x-1))/(x^2+x-2)
as x approaches 1

Answer 1

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

First off, we determine if it is of an indeterminate form. If we let
f(x) = sin(x - 1) / (x^2 + x - 2), then
f(1) = sin(1 - 1) / (1^2 + 1 - 2) = sin(0)/(0) = 0/0, so it is of the indeterminate form [0/0], and we apply L'Hospital's rule.

Recall that L'Hospital's rule works by taking derivative of the numerator and denominator, forming a new limit altogether. That is

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

Is the same as taking the derivative of the top and bottom:

lim [cos (x - 1) / (2x + 1)]
x -> 1

We can now solve the limit directly by plugging in the value x = 1.

cos(1 - 1) / (2(1) + 1) = cos(0) / (3) = 1/3

Answer 2

We have that (x^2 + x -2) = (x -1)(x + 2). Therefore, (sin(x-1))/(x^2+x-2) = sin(x-1))/(x -1) * (1/(x +2). As x -> 1, x - 1 => 0. We know lim (t -> 0) sin(t)/ t = 1. Therefore, (x -> 1) sin(x -1)/(x -1) = 1 and lim (sin(x-1))/(x^2+x-2 = 1 * 1/(1 + 2) = 1/3

Answer 3

you gonna substitute every x by 1
it will be 0\0 un specified
so u will factorize lim sin(x-1) \ (x-1)(x+2)
since sin(x-1) \ (x-1) =1
lim 1\(x+2)
substitute every x by 1
it will be = 1\3

Answer 4

apply the well-known property:
lim sin(x)/x = 1, as x -> 0

ans = 1/3

Answer 5

lim (sin(x-1))/(x^2+x-2)=0/0 for x=1

derivative (hospital rule)

lim cos(x-1)/(2x+1)
for x=1
cos(1-1)/(2+1)
=1/3
answer: 1/3

Answer 6

.0058177

iyiogrenci (the answer above this one) is wrong because he/she didn't use the quotient rule