evaluate without using L'Hopital theorem the following limit
lim x-->0 [(sin(x)-x)/(x-tan(x))]
the answer is 0.5 but I want to know the steps to calculate such a problem
To evaluate the limit lim x-->0 [(sin(x)-x)/(x-tan(x))] without using L'Hopital's theorem, we can simplify the expression by multiplying the numerator and denominator by the conjugate of x-tan(x), which is x+tan(x). This helps us eliminate the trigonometric terms in the expression.
Let's start by multiplying the numerator and denominator by the conjugate:
[(sin(x) - x)/(x - tan(x))] * [(x + tan(x))/(x + tan(x))]
Now we can simplify the expression:
= (sin(x)* (x + tan(x)) - x * (x + tan(x))) / [(x - tan(x)) * (x + tan(x))]
= (x * sin(x) + sin(x) * tan(x) - x^2 - x * tan(x)) / (x^2 - tan^2(x))
Notice that (x^2 - tan^2(x)) = x^2 - sin^2(x)/cos^2(x) = x^2 * cos^2(x) - sin^2(x).
Now we can rewrite the expression as:
= [x * (sin(x) + tan(x)) - (x^2 * cos^2(x) - sin^2(x))] / (x^2 * cos^2(x) - sin^2(x))
Next, we can simplify the expression further:
= [x * sin(x) + x * tan(x) - x^2 * cos^2(x) + sin^2(x)] / [x^2 * cos^2(x) - sin^2(x)]
Now, let's group the terms:
= [x * (sin(x) + tan(x)) - x^2 * cos^2(x) + sin^2(x)] / [x^2 * cos^2(x) - sin^2(x)]
Now, let's cancel out the common factors:
= [x * (sin(x) + tan(x)) - x^2 * cos^2(x) + sin^2(x)] / [x^2 * (cos^2(x) - 1) + sin^2(x)]
Since we know that cos^2(x) - 1 = -sin^2(x), we can substitute this into the denominator:
= [x * (sin(x) + tan(x)) - x^2 * cos^2(x) + sin^2(x)] / [-x^2 * sin^2(x) + sin^2(x)]
= [x * (sin(x) + tan(x)) - x^2 * cos^2(x) + sin^2(x)] / sin^2(x) * (1 - x^2)
Now, we can simplify the expression further by factoring out sin^2(x):
= [(x * (sin(x) + tan(x)) - x^2 * cos^2(x) + sin^2(x))] / [sin^2(x) * (1 - x^2)]
= [(x * (sin(x) + tan(x)) + sin^2(x) * (1 - cos^2(x)))] / [sin^2(x) * (1 - x^2)]
= [(x * (sin(x) + tan(x)) + sin^2(x) * sin^2(x))] / [sin^2(x) * (1 - x^2)]
= [x * (sin(x) + tan(x)) + sin^4(x)] / [sin^2(x) * (1 - x^2)]
Now, we can substitute x=0 into the expression:
= [(0 * (sin(0) + tan(0)) + sin^4(0))] / [sin^2(0) * (1 - 0^2)]
= [0 + 0] / [0 * (1 - 0)]
= 0 / 0
Since we have 0/0, this is an indeterminate form. To evaluate the limit further, we can apply L'Hopital's theorem.