How do I derive the oblique asymptote for $$y = \frac{x+2}{e^x - 2}$$
In case it's needed, the supposed answer is $$ y = -0.5x - 1$$
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Find slope with $\lim_\infty\dfrac{y}{x}$ first – Nosrati Sep 05 '17 at 01:59
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Hint: Just find the slop $$m=\lim_{x\to\pm\infty}\dfrac{y}{x}$$ which gives $m=-\dfrac12$ as $x\to-\infty$ and then find $$h=\lim_{x\to\pm\infty}y-mx=-1$$
Nosrati
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Hello, do you mind explaining what value of y did you use?
To me, as x approaches $-\infty$, $$y \to \frac{e^(-\infty)}{-\infty} \to 0$$.
– Ong Hai Xiang Sep 05 '17 at 07:01 -
@OngHaiXiang As $x\to+\infty$, $\lim_{x\to\pm\infty}\dfrac{y}{x}\to0$and doesn't make oblique asymptote. – Nosrati Sep 05 '17 at 08:08
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Your function has no an oblique asymptote because $$\lim_{x\rightarrow+\infty}\frac{x+2}{e^x-2}=0$$ and $$\lim_{x\rightarrow-\infty}\frac{x+2}{e^x-2}=+\infty$$
Michael Rozenberg
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