limit with greatest integer function How can I evaluate this limit?
$$\lim_{x\to-7} \frac{[x]^2+15[x]+56}{\sin(x+7)\sin(x+8)}$$ where $[x]$ denotes the greatest integer less than or equal to $x$
I could easily factor the numerator. But I cannot apply any of the standard limits due to $[x]$.
 A: On the right of $-7$ you have $[x]=-7$, while on the left $[x]=-8$. In any case you have
$$\frac{[x]^2 + 15[x]+56}{\mbox{something}} = \frac{([x]+7)([x]+8)}{\mbox{something}} = 0$$
so the function is identically $0$ in a neighbourhood of $-7$. This implies that the limit is $0$.
A: Since $[x]$ is constant between integers, we can divide this into two parts: the left hand limit and the right hand limit.
When $-7 < x < -6$ we have $[x] = -7$, so $$\lim_{x\to -7^{+}} \frac{[x]^2+15[x]+56}{\sin(x+7)\sin(x+8)} = \lim_{x\to -7^{+}} \frac{(-7)^2+15(-7)+56}{\sin(x+7)\sin(x+8)}$$ and the rest proceeds as a normal limit
When $-8 < x < -7$ then $[x]=-8$ and proceed similarly for $\lim_{x\to -7^-}$.
A: The way the question is phrased makes me wonder if you've understood the question, either the nature of limits or the nature of the greatest-integer function.  Your final sentence makes me suspect the latter.
If $x<-7$ then $\lfloor x\rfloor=-8$.
If $x>-7$ then $\lfloor x\rfloor= -7$.
It you need that explained to you you then you haven't understood the greatest-integer function and that would be what you need to look at.
In one case the numerator is $(-8)^2+15(-8)+56=0$ and in the other case it is $(-7)^2 + 15(-7)+56 = 0$.  Either way the fraction is equal to $0$ in small neighboorhoods about $-7$ unless the denominator is $0$.  And in small neighborhoods about $-7$, the denominator is $0$ only if $x=-7$, so essentially your asking for $\lim\limits_{x\to-7} 0$.
