show that the curve sqrt(x^2+4x) has two asymptotes: y=x+2 and y+-x-2.
Slant asymptotes are lines that the curve approaches as x --> +oo or x --> -oo. As x --> +oo or -oo, $\displaystyle (x+2)^2 - 4 \rightarrow (x+2)^2$ and the square root of this is $\displaystyle \pm (x+2)$. This is not a rigorous way of doing it. The mathematicians of MHF will pounce on this and make mince meat of it.
But I'm not sure you're at a level where a rigorous approach would make much sense, so you'll just have to make do with the one above.
Hi all. I was confused by this limit too. I am not quite sure what is wrong with this reasoning. I have a suspicion but I would like to see other thoughts.
For x > 0, sqrt(x^2 + 4x) is the same as (x^2 + 4x)/sqrt(x^2 + 4x) = (x + 4)/sqrt(1 + 4/x).
At this point, why can we just let 4/x ->0 as x -> inf and we get the limit tending to x + 4?
the graph of $y = \sqrt{x^2+4x}$ is the upper branch of a hyperbola with a horizontal transverse axis ...
$y^2 = x^2+4x$
$y^2+4 = (x+2)^2$
$(x+2)^2 - y^2 = 4$
note the standard form for a hyperbola is $\dfrac{(x-h)^2}{a^2} - \dfrac{(y-k)^2}{b^2} = 1$
$\dfrac{(x+2)^2}{2^2} - \dfrac{y^2}{2^2} = 1$ , a hyperbola centered at $(h,k) = (-2,0)$
Asymptotes have slope $m = \pm \dfrac{b}{a} = \pm \dfrac{2}{2} = \pm 1$
since the asymptotes pass through the hyperbola center ...
$y - 0 = -1(x + 2) \implies y = -x - 2$
and
$y-0 = 1(x+2) \implies y = x+2$
The first thing I notice, is that it is not rigorous to state $\displaystyle (x+2)^2-4 \to x+2$. Instead we should be saying something like$\displaystyle (x+2)^2-4 = (x+2)^2{(x+2)^2-4 \over (x+2)^2} = (x+2)^2\left(1 - {4 \over (x+2)^2}\right) \to (x+2)^2$Then we get that$\displaystyle y = \sqrt{x^2+4x} \to \sqrt{(x+2)^2} = \left|x+2\right|$
And the modulus signs are critical, because when $\displaystyle x < -2$ as it is when $\displaystyle x \to -\infty$, this gives us $\displaystyle y \to -(x+2) = -x-2$.
OK. I realise you meant that you have some different reasoning that you know to be wrong.
The problem comes in how $\displaystyle \sqrt{1+\frac4x} \to 1$. The (Maclaurin) series expansion of $\displaystyle \sqrt{1+y}$ about $\displaystyle y=0$ is $\displaystyle \sqrt{1+y} = 1 + \tfrac12 y + \ldots$ so$\displaystyle \begin{aligned}{x + 4 \over \sqrt{1 + \frac4x}} &= {(x+4)\sqrt{1-\frac4x} \over \sqrt{1+\frac4x}\sqrt{1-\frac4x}} \\ & \approx {(x+4)(1-\frac2x) \over 1 - \frac4x} = {x + 4 - 2 - \frac8x \over 1 - \frac4x} = {x+2 - \frac8x \over 1 -\frac4x} \\ &\approx x+2 \end{aligned}$for large $\displaystyle x$.
Alternatively using the Maclaurin series above and $\displaystyle {1 \over 1+y} = 1 - y + \ldots$, we have
$\displaystyle \begin{aligned}{x + 4 \over \sqrt{1 + \frac4x}} &\approx {x+4 \over 1 + \frac2x} \\ &\approx (x+4) (1 - \frac2x) = x + 4 - 2 - \frac8x \\ &\approx x+2\end{aligned}$