Find $\lim_{(x,y)\to (\alpha,0)}\left(1+\frac{x}{y}\right)^y$ Find $$L=\lim_{(x,y)\to (\alpha,0)}\left(1+\frac{x}{y}\right)^y$$
My try:
I have chosen the Path $x=my+\alpha$
We get:
$$L=\lim_{y \to 0}\left(1+m+\frac{\alpha}{y}\right)^y$$
Which is of form $\infty^0$ and by L'Hopital's Rule i got $L=1$
Also by Path $x=0$ we get:
$$L=1$$
Now how can we conclude $L=1$ for any path we choose?
 A: for $x> 0 \ \ y> 0 $ we get the following:
    $$ \left( 1+\dfrac{x}{y}\right) ^{y}=\exp\left( y\ln(1+\dfrac{x}{y})\right)=\exp\left( y\ln(x+y)-y\ln(y)\right)   $$
    we have the following inequality:
    $$\ln(x+y)\le x+y\Longrightarrow y\ln(x+y)\le y(x+y) $$ 
$$y\ln(y)\le y\ln(x+y)\Longrightarrow 0\leq y\ln(x+y)-y\ln(y)$$ thus
$$1\leq \exp\left( y\ln(x+y)-y\ln(y)\right)\leq \exp\left(  y(x+y)-y\ln(y)\right)  $$
and because  $\displaystyle \lim_{(x,y)\rightarrow(\alpha,0)} y(x+y)-y\ln(y)=0$ then we get the result that $\displaystyle \lim_{(x,y)\rightarrow(\alpha,0)}\exp\left( y\ln(x+y)-y\ln(y)\right)=\lim_{(x,y)\rightarrow(\alpha,0)} \left( 1+\dfrac{x}{y}\right) ^{y}=1 $
A: There is a problem with the expression $(1+\frac  xy)^{y}$ when  $1+\frac  xy$ is negative. So I will handle the case when the limit is taken through positive values of $y$ and $\alpha >0$. 
$\lim_{y \to 0} (1+m+\frac {\alpha} y)^{y}=\lim_{y \to 0}(1+m)^{y}(1+\frac {\alpha /(1+m)} y)^{y} )=(1)(1)=1$. But this does not prove that the limit exists and equals $1$.
Observe that $(1+\frac t y)^{y}=e^{y\ln (1+\frac  t y)} \to e^{0}=1$ for each fixed $t $. Apply this to $t=\alpha -\epsilon$ and $\alpha +\epsilon$ and use the squeeze Theorem to conclude that the given limit exists and equals $1$.
