# How to find the oblique asymptote of this function when $x\to-\infty$?

I posted a question earlier and got the answer for $f(x)=x-\sqrt{x^2+5}$ .

I would like to know the oblique asymptote for $f(x)=x-\sqrt{x^2+5x}$

The function has a horizontal asymptote $y=\dfrac{-5}{2}$ when $x\to+\infty$. How do I find the oblique asymptote of this function when $x\to-\infty$?

• – Michael Rozenberg Dec 30 '17 at 7:14
• Try rationalising, ie multiplying and dividing by conjugate – samjoe Dec 30 '17 at 7:15

Let the oblique asymptote be $y = ax + b$ with $a,b$ are real constants. The definition of the oblique asymptote means that $x - \sqrt{x^2+5x} - ax -b \to 0$ as $x \to -\infty$ where $a,b$ are to be determined a bit later. We see that $a = 2$ immediately, and $-x-\sqrt{x^2+5x} - b= - \left(x+\sqrt{x^2+5x} +b\right)= \dfrac{5x}{x-\sqrt{x^2+5x}}-b=\dfrac{5x}{x-|x|\sqrt{1+\dfrac{5}{x}}}-b\to 0$ when $x \to -\infty$. So $b = \dfrac{5}{2}$. Thus the oblique asymptote is: $y = 2x+\dfrac{5}{2}$.
The slope will be the limit of the derivative: $$\lim_{x\to-\infty}1-\frac {2x+5}{2\sqrt {x^2+5x}}=\lim_{x\to-\infty}1+\frac {2+5/x}{2\sqrt {1+5/x}}=2.$$ The change of sign is due to $\sqrt {x^2}=-x$ when $x <0$. To find the $x$-intercept, the difference with the line should go to $0$: \begin{align} 0&= \lim_{x\to-\infty}{x-\sqrt {x^2+5x}}-({2x+b})= \lim_{x\to-\infty}-x-\sqrt {x^2+5x}-b\\ \ \\ &=\lim_{x\to\infty}x-\sqrt {x^2-5x}-b =\frac52-b.\\ \ \\ \end{align} So the asymptote is $y=2x+\frac52$.