Chủ đề này có 4 trả lời

[ttjkbongdem]

$\lim_{x->1}(2-x)^{tan(\dfrac{\Pi.x }{2})}$

[Crystal]

$\lim_{x->1}(2-x)^{tan(\dfrac{\Pi.x }{2})}$

Đặt: $x = 1 + t \Rightarrow tg\dfrac{{\pi \left( {1 + t} \right)}}{2} = - \dfrac{{c{\text{os}}\dfrac{{\pi t}}{2}}}{{\sin \dfrac{{\pi t}}{2}}},x \to 1 \Rightarrow t \to 0$
Do đó: $$A = {\left( {2 - x} \right)^{tg\dfrac{{\pi x}}{2}}} = {\left( {1 - t} \right)^{ - \dfrac{{c{\text{os}}\dfrac{{\pi t}}{2}}}{{\sin \dfrac{{\pi t}}{2}}}}}$$
Lấy logarit tự nhiên hai vế, ta được: $$\ln A = - \dfrac{{c{\text{os}}\dfrac{{\pi t}}{2}}}{{\sin \dfrac{{\pi t}}{2}}}\ln \left( {1 - t} \right) = - \dfrac{{1 + o\left( t \right)}}{{\dfrac{{\pi t}}{2} + o\left( t \right)}}\left( { - t + o\left( t \right)} \right)$$
$$ = \dfrac{2}{\pi }\left( {t + o\left( t \right)} \right)\left( {1 + o\left( t \right)} \right)\dfrac{1}{{1 + o\left( t \right)}}$$
Suy ra: $$\mathop {\lim }\limits_{t \to 0} \ln A = \dfrac{2}{\pi } \Rightarrow \mathop {\lim }\limits_{t \to 0} A = \mathop {\lim }\limits_{x \to 1} {\left( {2 - x} \right)^{tg\dfrac{{\pi x}}{2}}} = \boxed{{e^{\dfrac{2}{\pi }}}}$$

[Crystal]

$\lim_{x->1}(2-x)^{tan(\dfrac{\Pi.x }{2})}$

Cách khác dùng quy tắc L'Hospital.

Ta có: $$\mathop {\lim }\limits_{x \to 1} {\left( {2 - x} \right)^{tg\dfrac{{\pi x}}{2}}} = \mathop {\lim }\limits_{x \to 1} {e^{tg\dfrac{{\pi x}}{2}\ln \left( {2 - x} \right)}} = {e^{{{\mathop {\lim }\limits_{x \to 1} }^{tg\dfrac{{\pi x}}{2}\ln \left( {2 - x} \right)}}}}$$
Lại có: $$\mathop {\lim }\limits_{x \to 1} tg\dfrac{{\pi x}}{2}\ln \left( {2 - x} \right) = \mathop {\lim }\limits_{x \to 1} \dfrac{{\ln \left( {2 - x} \right)}}{{\cot g\dfrac{{\pi x}}{2}}}\,\,\,\,\,\,\left( {\dfrac{0}{0}} \right)$$
Theo quy tắc L'Hospital suy ra: $$\mathop {\lim }\limits_{x \to 1} tg\dfrac{{\pi x}}{2}\ln \left( {2 - x} \right) = \mathop {\lim }\limits_{x \to 1} \dfrac{{{{\left[ {\ln \left( {2 - x} \right)} \right]}^\prime }}}{{{{\left( {\cot g\dfrac{{\pi x}}{2}} \right)}^\prime }}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{ - \dfrac{1}{{2 - x}}}}{{ - \dfrac{\pi }{2}{{\sin }^2}\left( {\dfrac{{\pi x}}{2}} \right)}}$$
$$ = \mathop {\lim }\limits_{x \to 1} \dfrac{{{{\sin }^2}\left( {\dfrac{{\pi x}}{2}} \right)}}{{\dfrac{\pi }{2}\left( {2 - x} \right)}} = \dfrac{2}{\pi }$$
Vậy $$\mathop {\lim }\limits_{x \to 1} {\left( {2 - x} \right)^{tg\dfrac{{\pi x}}{2}}} = \boxed{{e^{\dfrac{2}{\pi }}}}$$

[thaovisp]

Nhân đây bạn cho mình hỏi bài này. Tính $\mathop {\lim }\limits_{x \to 0} {x^{{x^x} - 1}}$

[Crystal]

Nhân đây bạn cho mình hỏi bài này. Tính $\mathop {\lim }\limits_{x \to 0} {x^{{x^x} - 1}}$

Mình làm thế này.

Ta có: $$\mathop {\lim }\limits_{x \to 0} {x^{{x^x} - 1}} = \mathop {\lim }\limits_{x \to 0} {e^{\left( {{x^x} - 1} \right)\ln x}} = {e^{\mathop {\lim }\limits_{x \to 0} }}^{\left( {{x^x} - 1} \right)\ln x}$$
Lại có: $$\mathop {\lim }\limits_{x \to 0} \left( {{x^x} - 1} \right)\ln x = \mathop {\lim }\limits_{x \to 0} \left( {{e^{x\ln x}} - 1} \right)\ln x = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{{e^{x\ln x}} - 1}}{{x\ln x}}} \right)x{\ln ^2}x$$
$$ = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{{e^{x\ln x}} - 1}}{{x\ln x}}} \right)\mathop {\lim }\limits_{x \to 0} \left( {x{{\ln }^2}x} \right) = \mathop {\lim }\limits_{x \to 0} \left( {x{{\ln }^2}x} \right) = \mathop {\lim }\limits_{x \to 0} \dfrac{{{{\ln }^2}x}}{{\dfrac{1}{x}}}\,\,\,\,\,\,\left( {\dfrac{\infty }{\infty }} \right)$$
Áp dụng quy tắc L'Hospital:
$$\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\ln }^2}x}}{{\dfrac{1}{x}}} = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{2\dfrac{{\ln x}}{x}}}{{ - \dfrac{1}{{{x^2}}}}}} \right) = - 2\mathop {\lim }\limits_{x \to 0} x\ln x = - 2\mathop {\lim }\limits_{x \to 0} \dfrac{{\ln x}}{{\dfrac{1}{x}}} = - 2\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\dfrac{1}{x}}}{{ - \dfrac{1}{{{x^2}}}}}} \right)$$
$$ = 2\mathop {\lim }\limits_{x \to 0} x = 0 \Rightarrow \mathop {\lim }\limits_{x \to 0} \left( {{x^x} - 1} \right)\ln x = 0$$
Vậy $$\mathop {\lim }\limits_{x \to 0} {x^{{x^x} - 1}}\boxed{ = {e^0} = 1}$$
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Ngoài ra cũng có thể đặt $y = {x^{{x^x} - 1}}$ rồi lấy logarit tự nhiên hai vế ta cũng được kết quả.