Determine the limit of the trigonometric function:
Limit as x approaches pi/4 of (1-tanx)/(sinx-cosx)
I need step by step help using the special trignometric limits theorem.
(1-tanx)/(sinx-cosx) = 0/0, so try LHospital's Rule:
-sec^2(x)/(cosx+sinx) = (-2)/(2/√2) = -2/√2 = -√2
To determine the limit of the given trigonometric function, we can use the special trigonometric limits theorem. The theorem states that:
lim(x→a) sin(x) = sin(a), lim(x→a) cos(x) = cos(a), and lim(x→a) tan(x) = tan(a)
Given the function:
f(x) = (1 - tan(x))/(sin(x) - cos(x))
Let's evaluate the limit as x approaches pi/4 step by step:
Step 1: Simplify the expression using the special trigonometric limits theorem:
lim(x→π/4) (1 - tan(x))/(sin(x) - cos(x))
= (1 - tan(π/4))/(sin(π/4) - cos(π/4))
Step 2: Evaluate the trigonometric functions:
tan(π/4) = 1 (special limit theorem)
sin(π/4) = cos(π/4) = √2/2 (special limit theorem)
Step 3: Substitute the values into the expression:
(1 - 1)/(√2/2 - √2/2)
= 0/0
At this point, we have obtained an indeterminate form of 0/0, which means we can apply L'Hôpital's rule.
Step 4: Apply L'Hôpital's rule by differentiating the numerator and the denominator:
lim(x→π/4) (1 - tan(x))/(sin(x) - cos(x))
= lim(x→π/4) (1 - sec^2(x))/(cos(x) + sin(x))
Step 5: Evaluate the trigonometric functions:
sec^2(π/4) = cos^2(π/4) = sin^2(π/4) = 2
Step 6: Substitute the values into the expression:
(1 - 2)/(√2/2 + √2/2)
= -1/√2
Therefore, the limit as x approaches pi/4 of (1-tanx)/(sinx-cosx) is -1/√2.