Limit of f(x,y) as (x,y) approaches the origin point What is the limit of $$\frac {(1+y)^z-1} {(y^2+z^2)}$$ as (y,z) approaches the origin?
I thought of using l'hopitals rule because $(1+y)^z$ approaches 1 so numerator approaches 0, so does the denominator.
$$\frac {z(1+y)^{z-1}} {(2z)}$$=$$\frac {z^2(1+y)^{z-2}} {2}$$ which is <0? how would i prove that it is smaller than 0 rigourously? also, I am not sure how to prove (1+y)^z approaches 1, do i just plug in the two points (0,0)? 
 A: I would say that the numerator approximately equals $yz$
i.e. Start with a binomial expansion
$(1+y)^z = 1 + zy + \frac {1}{2} z(z-1)y^2 + \cdots$ if $y,z$ are small we can ignore the tail.
And consider the limit  $\lim_\limits{(y,z)\to 0}\frac {yz}{y^2 + z^2}$ and that limit does not exist.
A: The problem with your approach is that the L'Hopital rule for multivariable functions is not applied the way you have done it. In fact, there is no well-known rule for multi variable functions , so you must create parametrized paths leading to 1 dimensional limits and then apply L'Hopital. But still, see here for a multivariable L'Hopital that is complicated to apply but can be done.

The limit does not exist.
Indeed, while going via the line $y = 0$ or $z = 0$ to the origin, the above expression evaluates to $0$, and hence has limit $0$ while travelling along these paths.
However, going via the line $y =z$ we get the expression $\lim_{x \to 0} \frac{(1+x)^x - 1}{2x^2}$. Now, we note that the input is infinitely differentiable near $0$, so one may take derivatives and hope to apply L'Hopital rule :
$$
\frac{(1+x)^x - 1}{2x^2} \xrightarrow{diff} \frac{(1+x)^x \left[\frac{x}{1+x} + \log(1+x)\right]}{4x} \\ \xrightarrow{diff} \frac{(1-x)^{x-2}[x^2+x+(1+x)^2\log^2(1+x)+2x(1+x)\log(1+x)+2]}{4}
$$
Where the last expression is continuous at $0$, so exists and equals $\frac 24 = \frac 12$, whence by L'Hopital the initial limit also exists and equals $\frac 12$. Since $\frac 12 \neq 0$, the limit is not going to exist.
A: There is no L'hopital rule for multivariable functions.
Consider that in a single variable function, the variable may approach a point from two sides: the left, or the right. L'hopital's rule works because of the continuity of the derivative from the left and the right.
On the other hand, in two dimensions, the variables can approach a point in an infinite number of directions. Which variable would you differentiate by anyway?
It seems to me that you have partially differentiated with respect to $z$, however if you have, it is not correct anyway. 
In this case as $y$ and $z$ tend to the same thing, you may replace them identically with another variable, say, $x$, and then use L'hopital rule.
