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$$ \lim \limits_{(x,y) \to (1,2)}\frac{xy-2x-y+2}{x^2 + y^2 - 2x -4y + 5} $$

I can't seem to find a way to factor this out so I was wondering since this comes to 0/0 can I use l'hopitals rule on this and if so would it I take the partial derivative instead of the regular derivative.

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  • $\begingroup$ I think you can just plug it in. $\endgroup$ – Jacob Wakem Mar 22 '17 at 6:55
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    $\begingroup$ No, he can't just plug in as then the denominator vanishes... $\endgroup$ – DonAntonio Mar 22 '17 at 7:29
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Try to reach the limit from different directions. Let's consider first $y=x^2+1$; your expression becomes $$ \frac{x^3-x^2-x+1}{x^4-x^2-2x+2} $$ and using de l'Hopital you get

\begin{align*} \lim_{x\to1}\frac{x^3-x^2-x+1}{x^4-x^2-2x+2} &\stackrel{H}{=}\lim_{x\to1}\frac{3x^2-2x-1}{4x^3-2x-2}\\ &\stackrel{H}{=}\lim_{x\to1}\frac{6x-2}{12x^2-2}=2/5 \end{align*}

Next consider the direction $y=x+1$; now your expression becomes $$ \frac{x^2-2x+1}{2x^2-4x+2} $$ and again with de L'Hopital

\begin{align*} \lim_{x\to1}\frac{x^2-2x+1}{2x^2-4x+2} &\stackrel{H}{=}\lim_{x\to1}\frac{2x-2}{4x-4}\\ &\stackrel{H}{=}\lim_{x\to1}\frac{2}{4}=1/2 \end{align*}

Thus we have found two different limits, which is absurd. Thus your function doesn't admit limit.

Another way: notice that

\begin{align*} \frac{xy-2x-y+2}{x^2+y^2-2x-4y+5} &=\frac{\overbrace{(x-1)}^{a}\overbrace{(y-2)}^{b}}{(x-1)^2+(y-2)^2}\\ &=\frac{ab}{a^2+b^2} \end{align*} thus your limit is equal to $$ \lim_{(a,b)\to(0,0)}\frac{ab}{a^2+b^2} $$ and again you can show that the limit doesn't exist as above. Consider $a=b$, and your expression becomes $$ \frac{a^2}{2a^2}\equiv1/2 $$

and then with $b=a^2$ our expressione equals $$ \frac{a^3}{a^2+a^4}=\frac{a}{1+a^2}\stackrel{a\to0}{\longrightarrow}0 $$ so, again you've found two different limits.

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At first, we change the form of limit by replacing $x=X+1$ and $y=Y+2$ that result that $$ \lim \limits_{(x,y) \to (1,2)}\frac{xy-2x-y+2}{x^2 + y^2 - 2x -4y + 5}= \lim \limits_{(X,Y) \to (0,0)}\frac{X\,Y}{X^2+Y^2} $$

Now, One of the clasical method for obtaining the limit of function with two variables in $(0,0)$ is by this fact that you can assume $X=r \, \cos(\theta)$ and $Y=r \, \sin(\theta)$, then calculate the limit with new values for $X$ and $Y$ when $r \to 0$, if after calculation limit, the final values of limit depend on to $\theta$, we conclude that, the function dose not have limit in $(0,0)$ but if the final values of limit be independent of $\theta$, we say the function has limit in $(0,0)$ with value that we obtained. so we have

$$\lim_{(X,Y) \to (0,0)}\,\dfrac { X\,Y } {X^{2}+Y^{2} } =\lim_{(r) \to (0)}\frac{r^2\,\cos(\theta)\, \sin(\theta)}{r^2}=\cos(\theta)\, \sin(\theta) $$ that means your limit does not exist in $(0,0)$.

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    $\begingroup$ I suppose you meant $y=Y+2$ $\endgroup$ – Claude Leibovici Mar 22 '17 at 8:15
  • $\begingroup$ @ClaudeLeibovici Oh yahh. Thanks for comment. I edit it $\endgroup$ – Amin235 Mar 22 '17 at 8:16
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As $(x,y)$ approaches $(1,2)$ along the line $y=2x$ the value of the function is constantly $2/5.$

As $(x,y)$ approaches $(1,2)$ along the line $x=1$ the numerator is $0$ and the denominator is $(2-y)^2\ne 0$ so the value of the function is constantly $0$.

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