Question

(10 marks) It can be shown that the one-sided limit lim x→0 + ​ (1−tan2t) 1/t is a finite number. Consequently, (1−tan2t) 1/t is continuous and integrable on [0,1]. Use this to evaluate lim x→0 ​ x 1 ​ ∫ 0 x ​ (1−tan2t) 1/t dt.

Answer 1

To evaluate the limit involving an integral, one applies the Fundamental Theorem of Calculus and potentially L'Hôpital's Rule to handle indeterminate forms. The given function is continuous and integrable on the interval [0,1], enabling the application of these calculus principles.

Step-by-step explanation:

The question asks to evaluate the limit lim x→0 ​ x 1 ​ ∫ 0 x ​ (1−tan2t) 1/t dt. Given that the one-sided limit is finite and the function (1−tan2t) 1/t is continuous and integrable on [0,1], we can apply the Fundamental Theorem of Calculus to evaluate the limit.

To find this limit, we consider the definition of the definite integral as the accumulated area under the curve and relate it to the concept of continuity and differentiability. By the Fundamental Theorem of Calculus, if F(x) is the antiderivative of f(t), then ∫ 0 x f(t) dt = F(x) - F(0), and the derivative of F(x) with respect to x is f(x).

Applying L'Hôpital's Rule to deal with the indeterminate form 0/0, we differentiate the numerator and denominator of the fraction separately with respect to x. Since we're interested in the limit as x approaches 0 from the right, we then substitute the values and compute the limit. If this process results in a finite number, it concludes the evaluation of the limit as x approaches 0.