Tính giới hạn sau: \(\mathop {\lim }\limits_{x \to 0} {(\cos x)^{1/{x^2}}}\)
Trả lời:
Đáp án đúng: D
To evaluate the limit \(\mathop {\lim }\limits_{x \to 0} {(\cos x)^{1/{x^2}}}\), we use the logarithm method. Let \(y = {(\cos x)^{1/{x^2}}}\). Then, \(\ln y = \frac{1}{{{x^2}}}\ln (\cos x)\). We have \(\mathop {\lim }\limits_{x \to 0} \ln y = \mathop {\lim }\limits_{x \to 0} \frac{{\ln (\cos x)}}{{{x^2}}}\). This is in the indeterminate form \(\frac{0}{0}\), so we can apply L'Hopital's rule: \(\mathop {\lim }\limits_{x \to 0} \frac{{\ln (\cos x)}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{ - \sin x}}{{\cos x}}}}{{2x}} = \mathop {\lim }\limits_{x \to 0} \frac{{ - \tan x}}{{2x}}\) Applying L'Hopital's rule again (or using the fundamental limit \(\mathop {\lim }\limits_{x \to 0} \frac{{\tan x}}{x} = 1\)): \(\mathop {\lim }\limits_{x \to 0} \frac{{ - \tan x}}{{2x}} = - \frac{1}{2}\mathop {\lim }\limits_{x \to 0} \frac{{\tan x}}{x} = - \frac{1}{2} \cdot 1 = - \frac{1}{2}\) Thus, \(\mathop {\lim }\limits_{x \to 0} \ln y = - \frac{1}{2}\). Therefore, \(\mathop {\lim }\limits_{x \to 0} y = {e^{ - 1/2}}\) Hence, \(\mathop {\lim }\limits_{x \to 0} {(\cos x)^{1/{x^2}}} = {e^{ - 1/2}}\)
Bộ câu hỏi trắc nghiệm ôn thi môn Toán cao cấp A1 có đáp án dành cho các bạn sinh viên Đại học - Cao đẳng ôn thi dễ dàng hơn. Mời các bạn cùng tham khảo!
30 câu hỏi 60 phút
