left and right hand limits
I have this limit: $$ \lim_{x\to2}\frac{x^2}{2x-4}$$ and I know (by looking at Wolfram Alpha) that this function does not have a limit, but it has left and right hand limits. Firstly, how do I determine that the limit does not exists? And second, how do I calculate the left and right hand limits?
$\endgroup$ 02 Answers
$\begingroup$To begin, note that the limit will exist if and only if the left hand and right hand limits both exist and agree with each other.
Let us think informally about the behavior of the function as $x \rightarrow 2$ from either side. Approaching from the right, we see the numerator is approaching $4$ whereas the denominator is approaching $0$ (think: getting arbitrarily small). At the same time, the whole fraction is always positive. So what is $\displaystyle \lim_{x \rightarrow 2^+} \frac{x^2}{2x - 4}$?
If we instead approach from the left, once again the numerator approaches $4$ and the denominator approaches $0$. However, this time the fraction is always negative since $2x - 4 < 0$ when $x < 2$. So what is $\displaystyle \lim_{x \rightarrow 2^-} \frac{x^2}{2x - 4}$?
If you're feeling shaky with the above reasoning, I encourage you to plug, say, $x = 1.9$ and $x = 1.99$ into the fraction to get a more concrete sense of what is happening when approaching from the left, and likewise $x = 2.1$ and $x = 2.01$ when approaching from the right. If desired, there is no shame in doing this sort of experimentation. Once you have the basic intuition, you can add as much rigor as you'd like with an epsilon-delta type argument.
$\endgroup$ 6 $\begingroup$Let $x=2+h$, so that when $x\to2$, $h\to0$ $$\lim_{x\to2^+}\frac{x^2}{2x-4}=\lim_{h\to0^+}\frac{4+h^2+4h}{2h}=\lim_{h\to0^+}\left(\frac2h+\frac h2+2\right)=\infty$$ $$\lim_{x\to2^-}\frac{x^2}{2x-4}=\lim_{h\to0^-}\frac{4+h^2+4h}{2h}=\lim_{h\to0^-}\left(\frac2h+\frac h2+2\right)=-\infty$$
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