Vertical asymptote of $h(x)=\frac{x^2e^x}x$ $$h(x)=\frac{x^2e^x}x$$
The function h is defined above. Which of the following are true about the graph of $y=h(x)$?


*

*The graph has a vertical asymptote at $x=0$

*The graph has a horizontal asymptote at $y=0$

*The graph has a minimum point



A. None
B. 1 and 2 Only
C. 1 and 3 Only
D. 2 and 3 Only
E. 1, 2, and 3

I thought it was E but the answer is D, meaning the vertical asymptote is not at $x=0$, why? If you set the denominator to $0$ it equals $0$ since $x=0$.
 A: The denominator equaling zero is not enough to make a vertical asymptote. You get a vertical asymptote in the ratio of two continuous functions if the denominator is zero and the numerator is not zero.
In your particular case your function is the same as $xe^x$ near zero. The graph has a "hole" in it, also called a removable discontinuity.  You do not get an asymptote since a factor of $x$ cancels in numerator and denominator. In general, if both numerator and denominator are zero you may or may not get an asymptote.
A: The function, as you've written it, appears to be the same as
$f(x) = x e^x$ (apart from right at zero)
You can easily see the behavior by plotting this function.
Plot this at www.fooplot.com/ using    $x \cdot 2.718^x$
A: If you are allowed negative values of $x$, then you do get a horizontal asymptote at $y = 0$ for negative $x$ because $x e^x$ increases up to $0$ (but never reaches it) as $x$ becomes larger and larger magnitude negative numbers. Also, if you are allowed negative $x$, then the function has a minimum. The derivative is $(x+1)e^x$ which is $0$ at $x = -1$ which is where the minimum is achieved. So if negative $x$ is allowed, then 2 and 3 are true.
