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I have the function: $f(x)=2x-2^{x}+2$

I know that this function has an oblique asymptote, but all the tutorials I can find on google, are with rational functions with the form:

$$f(x)=\frac{P(x)}{Q(x)}$$

Where they simply just divide the denominator with the numerator.

But I can't do that, because my equation doesn't contain any fractions. So my question is: how do I find the function to the oblique asymptote for my $f(x)$?

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An asymptote is a line that approximates the function as $x\rightarrow\pm\infty$. In other words, you want a function of the form $f(x)=ax+b$ such that \begin{equation} \lim_{x\rightarrow\pm\infty}\frac{2x-2^x+2}{ax+b}=1 \end{equation} This is not possible for $x\rightarrow +\infty$ since the $2^x$ explodes, so you could have an asymptote only for $x\rightarrow -\infty$: \begin{equation} \lim_{x\rightarrow-\infty}\frac{2x-2^x+2}{ax+b}=\lim_{x\rightarrow-\infty}\left(\frac{2x+2}{ax+b}-\frac{2^x}{ax+b}\right)=1 \end{equation} Since $\lim_{x\rightarrow -\infty}\frac{2^x}{ax+b}=0$, you are left with \begin{equation} \lim_{x\rightarrow -\infty}\frac{2x+2}{ax+b}=1 \end{equation} which yelds $a=2$, $b=2$

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  • $\begingroup$ Thanks man, exactly what I needed! :) $\endgroup$ – MapleGuy Sep 14 '14 at 21:11
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    $\begingroup$ :) Actually Mosk' answer below is a bit more formal and general! $\endgroup$ – marco trevi Sep 15 '14 at 6:53
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$$\text {Hint: for}\ {x\to -\infty}, \ {2^x\to0}$$ $$\text{the oblique asymptote has the form }\ y=mx+q$$ $$\text{and to find m and q you calculate the following limits:}$$ $$\ m=\lim_{x\to-\infty}\ f(x)/x$$ $$\ q=\lim_{x\to-\infty}\ [f(x)-mx]$$

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