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Question Number 101096 by student work last updated on 30/Jun/20

find the oblique asymptote of f (x) = x \cdot e^{\frac{1}{x}}

i need your help

Commented by student work last updated on 30/Jun/20

i need soon

Commented by student work last updated on 30/Jun/20

who can help me ?

Answered by mathmax by abdo last updated on 30/Jun/20

\lim_{x \to + \infty} f (x) = \lim_{x \to + \infty} {xe}^{\frac{1}{x}} = \lim_{x \to + \infty} x = + \infty

\lim_{x \to + \infty} \frac{f (x)}{x} = \lim_{x \to + \infty} e^{\frac{1}{x}} = 1

\lim_{x \to + \infty} f (x) - x = \lim_{x \to + \infty} x e^{\frac{1}{x}} - x = \lim_{x \to + \infty} x (e^{\frac{1}{x}} - 1) but

e^{\frac{1}{x}} = 1 + \frac{1}{x} + o (\frac{1}{x^{2}}) \Rightarrow e^{\frac{1}{x}} - 1 = \frac{1}{x} + o (\frac{1}{x^{2}}) \Rightarrow x (e^{\frac{1}{x}} - 1) = 1 + o (\frac{1}{x}) \Rightarrow

\lim_{x \to + \infty} f (x) - x = 1 \Rightarrow the line y = x + 1 is oblique assymptote to C_{f} at + \infty

same result at - \infty