The following question asks me to find all horizontal, vertical and inclined asymptotes for the following function:
$f(x)=\sqrt{x^2+5x+1}$
I've learnt that to find vertical asymptotes, you let the denominator equal to zero. For horizontal asymptotes, you divide the x's top and bottom with the highest degree. To find inclined or slanted asymptotes if $\displaystyle\lim_{x\to\infty}[f(x)-(mx+c)]=0$ or $\displaystyle\lim_{x\to-\infty}[f(x)-(mx+c)]=0$.
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Hint: Show that $$\lim_{x\to \infty}\left(\sqrt{x^2+5x+1}-(x+\frac{5}{2})\right)=0.$$
As $x\to-\infty$, something similar happens, but we have to be very careful about sign. The approximating linear function is now $-x-\frac{5}{2}$.
For me, the only safe way to deal with $x$ large negative is to let $t=-x$, find out what happens when $t$ is large positive, and translate back.
Remark: You are probably expected to prove that the limit is $0$. It can be done in various ways, in particular by the "rationalizing the numerator" trick you may have used before.
The intuition is that if $x$ is large positive, then $x^2+5x+1$ is roughly of size $x^2$, but much closer to $(x+\frac{5}{2})^2$.
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Actually for the horizontal asymptote, don't worry you didn't answer your own question. If you'd been given a rational function, yes you would divide the highest power of x on top by highest power of x on bottom. But your function isn't even rational. It's just a square root, and there's actually no horizontal asymptote for it because its y value keeps increasing without bound as x increases. If you want more help on how to find the horizontal asymptote, here's a video that explains and gives examples: http://www.youtube.com/watch?v=0cPptjKTR7M
Good luck!
We know that if at least one of the following conditions: $$\lim_{x\to x_0^+}f(x)=\infty,~~\lim_{x\to x_0^-}f(x)=\infty$$ holds, the line $x=x_0$ would be the vertical asymptote and you see that because of the structure of $f(x)$ here, it doesn't have such this line. Also, if at least one of the conditions $$\lim_{x\to +\infty}f(x)=y_0,~~\lim_{x\to -\infty}f(x)=y_0$$ holds so the line $y=y_0$ is the horizontal asymptote, but when $x\to\infty$ we have $$f(x)\sim\sqrt{x^2}=|x|$$ This shows that we don't have any horizontal asymptotes for $f(x)$'s graph. Now think about the following limits: $$m=\lim_{x\to +\infty}\frac{f(x)}{x}=\lim_{x\to +\infty}\frac{|x|}{x}=1$$$b=\lim_{x\to +\infty}[f(x)-mx]=\lim_{x\to +\infty}[\sqrt{x^2+5x+1}-x]=\lim_{x\to +\infty}\left(\frac{(\sqrt{x^2+5x+1}-x)(\sqrt{x^2+5x+1}+x)}{(\sqrt{x^2+5x+1}+x)}\right)=\lim_{x\to +\infty}\left(\frac{(\sqrt{x^2+5x+1})^2-x^2}{(\sqrt{x^2+5x+1}+x)}\right)=\lim_{x\to +\infty}\left(\frac{+5x+1}{|x|\sqrt{1+5/x^2+1/x^2}+x}\right)=\lim_{x\to +\infty}\left(\frac{x(5+1/x)}{2x}\right)=5/2$ We you could find these constants $m$ and $b$, then the line $mx+b$ would be the inclined asymptote. Of course the same lmits may be taken for $m$ and $b$ while $x\to -\infty$.