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I can't resolve limit for oblique asymptote for: $$\frac{2x^2 e^{1/x}}{2x+1}-x$$ I've tried solving it by putting the common denominator but it's a bit confusing because the numerator is bigger then the denominator....

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    $\begingroup$ Are you sure the function is not only the fraction? $\endgroup$ – Bernard Aug 20 '15 at 19:01
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$\displaystyle\lim_{x\to\infty}\left(\frac{2x^2 e^{1/x}}{2x+1}-x\right)=\lim_{x\to\infty}\frac{2x^2e^{1/x}-2x^2-x}{2x+1}=\lim_{x\to\infty}\frac{2xe^{1/x}-2x-1}{2+\frac{1}{x}}=\lim_{x\to\infty}\frac{2x(e^{1/x}-1)-1}{2+\frac{1}{x}}=\frac{1}{2}$

since $\displaystyle\lim_{x\to\infty}x(e^{1/x}-1)=\lim_{x\to\infty}\frac{e^{1/x}-1}{\frac{1}{x}}=\lim_{t\to0^{+}}\frac{e^t-1}{t}=\lim_{t\to0^{+}}\frac{e^t}{1}=1$ using L'Hospital's rule.

Similarly, $\displaystyle\lim_{x\to -\infty}\left(\frac{2x^2 e^{1/x}}{2x+1}-x\right)=\frac{1}{2}$, so $y=\frac{1}{2}$ is the only horizontal asymptote for the graph,

and there is no oblique asymptote.

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There is no oblique asymptote here as it does not pass standard tests (verify them). Only a vertical asymptote at $ x=-1/2 $ exists due to zero of denominator.

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