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I have the following bivariate function where $\alpha,\beta,\gamma,\delta$ are constants, $a,b,c,d$ are integers and $x,y\in [0,1]$ $$ f(x,y) = \frac{\alpha(1-x^ay^b) + \beta(1-x^cy^d)}{\gamma(1-x^2)+\delta(1-y^3)} $$ which I would like to examine the limit of as $x=y=1$ from below. My instinct was to apply l'Hospital's rule by setting $x=y$ and converting to a univariate function. $$ f(x,x) = \frac{\alpha(1-x^{(a+b)}) + \beta(1-x^{(c+d)})}{\gamma(1-x^2)+\delta(1-x^3)} $$ from which we can gather the limit is $$ \lim_{x\rightarrow 1}f(x,x) = \frac{\alpha(a+b)+\beta(c+d)}{2\gamma +3\delta} $$ However, am also aware that l'Hospital's rule is strictly for univariate functions. In this particular case, how can I be certain that my limit is correct?

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    $\begingroup$ Looking at your result, I believe you have a typo in the definition of $f(x,y)$: the last $y^b$ in the numerator should be $y^d$. $\endgroup$ Feb 5 at 13:59
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Your approach is incomplete. If the bidimensional limit exists, the univariate one that you presented also exists and has the same value. However, the computation of the univariate limit only shows that if the bidimensional exists, it must have the same value.

What you have computed, by taking $y=x$, is called a directional limit. You could have considered, more generally, that $y = 1 + m(x-1)$. In this case, the directional limit (here you may use L'Hopital's rule) is given by $$ \lim_{x\to 1} f(x, 1+m(x-1))= \frac{\alpha (a + b m) + \beta (c + d m)}{3 \delta m + 2 \gamma}. $$

The dependence in $m$ proves that the bidimensional limit does not exist.

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If you set $y=1$, the expression turns to

$$\lim_{x\to 1}\frac{\alpha(1-x^a)+\beta(1-x^c)}{\gamma(1-x^2)}$$ and by L'Hospital,

$$=\lim_{x\to 1}\frac{\alpha ax^{a-1}+\beta cx^{c-1}}{\gamma2x}=\frac{\alpha a+\beta c}{\gamma2}.$$

But with $x=1$, you will get a very different expression.

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If $g,\,h$ are each $0$ at $x=y=1$ and can be expanded about it as Taylor series,$$g=g_x(1,\,1)(x-1)+g_y(1,\,1)(y-1)+o(x-1)+o(y-1)$$etc., but the limiting behaviour of $g/h$ is path-dependent unless$$\Delta:=g_x(1,\,1)h_y(1,\,1)-g_y(1,\,1)h_x(1,\,1)=0.$$With $g:=\alpha+\beta-\alpha x^ay^b-\beta x^cy^d,\,h=\gamma+\delta-\gamma x^2-\delta y^3$, we get$$\Delta=(b\alpha+d\beta)2\gamma-(a\alpha+c\beta)3\delta,$$so in general the limit is undefined.

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