Recognize singularity: We are given the limit to evaluate: limx→0(x1−cot(x))First, we recognize that both x1 and cot(x) have singularities at x=0, which means we need to find a way to combine these terms to eliminate the singularity. We can do this by finding a common denominator.
Find common denominator: To find a common denominator, we need to express cot(x) in terms of sine and cosine:cot(x)=sin(x)cos(x)Now we can rewrite the limit as:x→0lim(x1−sin(x)cos(x))
Combine fractions: The common denominator between x1 and sin(x)cos(x) is xsin(x). We multiply the numerator and denominator of x1 by sin(x) to get:x→0lim(xsin(x)sin(x)−sin(x)cos(x))
Apply L'Hôpital's Rule: Now that we have a common denominator, we can combine the fractions: limx→0(xsin(x)sin(x)−xcos(x))
Differentiate numerator and denominator: We can now apply L'Hôpital's Rule because the limit is in an indeterminate form (0/0). We differentiate the numerator and the denominator with respect to x: Numerator's derivative: dxd(sin(x)−xcos(x))=cos(x)−(cos(x)−xsin(x)) Denominator's derivative: dxd(xsin(x))=sin(x)+xcos(x)
Evaluate limit: We simplify the derivatives:Numerator's derivative simplifies to: cos(x)−cos(x)+xsin(x)=xsin(x)Denominator's derivative is already simplified: sin(x)+xcos(x)Now we can rewrite the limit using these derivatives:limx→0(sin(x)+xcos(x)xsin(x))
Apply L'Hôpital's Rule again: We can now evaluate the limit by substituting x=0 into the simplified expression: x→0lim(sin(x)+xcos(x)xsin(x))=sin(0)+0cos(0)0sin(0)=00 However, this is still an indeterminate form, which means we need to apply L'Hôpital's Rule again.
Differentiate second time: We differentiate the numerator and denominator again:Numerator's second derivative: dxd(xsin(x))=sin(x)+xcos(x)Denominator's second derivative: dxd(sin(x)+xcos(x))=cos(x)+cos(x)−xsin(x)=2cos(x)−xsin(x)
Evaluate final limit: We simplify the second derivatives:Numerator's second derivative simplifies to: sin(x)+xcos(x)Denominator's second derivative simplifies to: 2cos(x)−xsin(x)Now we can rewrite the limit using these second derivatives:limx→0(2cos(x)−xsin(x)sin(x)+xcos(x))
Evaluate final limit: We simplify the second derivatives:Numerator's second derivative simplifies to: sin(x)+xcos(x)Denominator's second derivative simplifies to: 2cos(x)−xsin(x)Now we can rewrite the limit using these second derivatives:limx→0(2cos(x)−xsin(x)sin(x)+xcos(x))We can now evaluate the limit by substituting x=0 into the second derivatives:limx→0(2cos(x)−xsin(x)sin(x)+xcos(x))=2cos(0)−0sin(0)sin(0)+0cos(0)=20=0
More problems from Evaluate definite integrals using the power rule