limit as t approaches infinity of te^{1/t}-t

解答

t \to \infty lim (t e^{\frac{1}{t}} - t)

1

求解步骤

t \to \infty lim (t e^{\frac{1}{t}} - t)

乘以

t e^{\frac{1}{t}} - t

的共轭根式

: \frac{e ^{\frac{2}{t}} t ^{2} - t ^{2}}{t e ^{\frac{1}{t}} + t}

= t \to \infty lim (\frac{e ^{\frac{2}{t}} t ^{2} - t ^{2}}{t e ^{\frac{1}{t}} + t})

化简

\frac{e ^{\frac{2}{t}} t ^{2} - t ^{2}}{t e ^{\frac{1}{t}} + t} : \frac{t ( e ^{\frac{2}{t}} - 1 )}{e ^{\frac{1}{t}} + 1}

= t \to \infty lim \frac{t ( e ^{\frac{2}{t}} - 1 )}{e ^{\frac{1}{t}} + 1}

x \to a lim [\frac{f ( x )}{g ( x )}] = \frac{lim _{x \to a} f ( x )}{lim _{x \to a} g ( x )}, x \to a lim g (x) \neq = 0

不定型除外

= \frac{lim _{t \to \infty} ( t ( e ^{\frac{2}{t}} - 1 ) )}{lim _{t \to \infty} ( e ^{\frac{1}{t}} + 1 )}

t \to \infty lim (t (e^{\frac{2}{t}} - 1)) = 2

t \to \infty lim (e^{\frac{1}{t}} + 1) = 2

= \frac{2}{2}

化简

= 1

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x \to \infty lim (7 + \frac{8}{x})