Vtyjkyuk

I want to find the asymptote oblique of the following function for $xrightarrowpminfty$

$$f(x)=sqrtx^2+3x=sqrtx^2left(1+frac3xx^2right)simsqrtx^2=|x|$$

For $xrightarrow+infty$ we have:

$$fracf(x)xsimfracx=fracxx=1$$

which means that the function grows linearly.

$$f(x)-mx=sqrtx^2+3x-x=xleft(sqrtfracx^2x^2+frac3xx^2-1right)sim xleft(frac 12cdotfrac3xright)=frac32$$

The oblique asymptote is $y=x+frac 3 2$ which is correct. For $xrightarrow-infty$ we have:

$$fracf(x)x=fracx=frac-xx=-1$$

This means that

$$f(x)-mx=sqrtx^2+3x+x=xleft(sqrtfracx^2x^2+frac3xx^2+1right)=xleft(sqrt1+frac3x+1right)sim xcdot2rightarrow -infty$$

Which is not what my textbook reports ($-frac32$). Any hints on what I did wrong to find the $q$ for $xrightarrow-infty$?

The problem is in
$$
f(x)-mx=sqrtx^2+3x+xundersetbeginarrayc uparrow \ textproblemendarray=xleft(sqrtfracx^2x^2+frac3xx^2+1right)=xleft(sqrt1+frac3x+1right)sim xcdot2rightarrow -infty
$$
which should be
$$
f(x)-mx=sqrtx^2+3x+x=xleft(-sqrtfracx^2x^2+frac3xx^2+1right)
$$
because we are near $-infty$, so $sqrtx^2=-x$.

In order to avoid such common mistakes, I suggest to change $x=-t$, when the limit for $xto-infty$ is involved:
$$
lim_xto-inftybigl(sqrtx^2+3x+xbigr)=
lim_ttoinftybigl(sqrtt^2-3t-tbigr)=
lim_ttoinftyfrac-3tsqrtt^2-3t+t
$$
From here it should be clear.

Another strategy, in this case, is to set $x=-1/t$, that transforms the limit into
$$
lim_tto0^+left(sqrtfrac1t^2-frac3t-frac1tright)=
lim_tto0^+fracsqrt1-3t-1t
$$
which is a simple derivative: if $f(t)=sqrt1-3t$, then $f'(t)=frac-32sqrt1-3t$ and
$$
lim_tto0^+fracsqrt1-3t-1t=f'(0)=-frac32
$$

This is not a different way, but is better I think
$$f(x)=sqrtx^2+3x=sqrtx^2+3x+(frac32)^2-(frac32)^2=sqrtleft(x+frac32right)^2-(frac32)^2simleft|x+frac32right|$$
which gives both oblique asymptotes.

$$f(x)=sqrtx^2+3x=|x|left(1+3over xright)^1over2$$ which, as $xrightarrow -infty$, is equal to $$-xleft(1-3over2xright)^1over2 sim -xleft(1+3over2xright)= -x -3over2$$ using Taylor or Bernoulli