Determine the limit of $\frac{3-x^2}{6-x}$ as $x$ tends to 6 from the left.

Now I know the limit the is negative infinity but how do I prove this?

I'm a bit unsure as to whether I can use certain intuitions as rules. For example I argued that for $x$ between $5$ and $6$, $\frac{3-x^2}{6-x}<\frac{-22}{6-x}$. Then since $6-x$ tends to $0$ from the right, I can say the limit is negative infinity. So by comparison, the original limit is the same

But am I right in thinking the quotient rule of limits holds when the denominator limit is infinity and the numerator is constant?

  • $\begingroup$ You can use the fact that $\lim_{x\to0^+}\frac{1}{x}=+\infty$ changing $x$ with $6-x$ and $x$ approach $6^-$ and replacing $1$ with $-22$ you get $-\infty$ $\endgroup$ – kingW3 Oct 30 '14 at 11:29

$\forall M\in \mathbb{N}$, take $\delta=\min(1,\frac{22}{M})>0$, then $\forall x$,s.t $0<6-x<\delta$, $$\frac{3-x^2}{6-x}<\frac{-22}{6-x}<-M$$

Hence $\lim_{x\to 6^-}\frac{3-x^2}{6-x}=-\infty$.


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