# The limit of $(1+x^2+y^2)^{1\over x^2 + y^2 +xy^2}$ as $(x,y)$ approaches $(0.0)$ [duplicate]

Im having trouble solving the following limit problem:

Whats the limit of $$(1+x^2+y^2)^{1\over x^2 + y^2 +xy^2}$$ as $$(x,y)$$ approaches $$(0.0)$$.

I know that the first step is to change $$(x,y)$$ to polar coordinates. However after I've done that and simplified the expression I'm left with the following: $$(1+r^2)^{1\over r^2+r^2\cos(α) \sin^2(α)}.$$ How do i prove that the expression goes to $$e$$ when $$r→0$$?

• Is this $$(1+x^2+y^2)^{\frac{1}{x^2+y^2+xy^2}}$$? – Dr. Sonnhard Graubner Jul 13 '19 at 8:50
• Yes, i'm so sorry for the poor formatting. – Fosorf Jul 13 '19 at 8:51
• Also, exact duplicate to Limit as (x,y) approaches (0,0) of ... – Lutz Lehmann Jul 13 '19 at 9:13
• A suggestion: You could rewrite your equation to $(1+r^2) ^ {\frac{1}{r^2}}$ and then substitute $r^2$ with $r^2 = \frac{1}{v}$ and you get the definition of $e$. – Imago Jul 13 '19 at 9:24
• MMA says the result is given by $$\sqrt{e}$$ – Dr. Sonnhard Graubner Jul 13 '19 at 10:07

You can write it using the limit laws, esp. the one on composition of continuous functions, as $$\lim_{(x,y)\to (0,0)}(1+x^2+y^2)^{1\over x^2+y^2+xy^2}=\lim_{(x,y)\to (0,0)}\left(\lim_{(x,y)\to (0,0)}(1+x^2+y^2)^{1\over x^2+y^2}\right)^{x^2+y^2 \over x^2+y^2+xy^2} \\ =\lim_{(x,y)\to (0,0)} \exp\left({x^2+y^2 \over x^2+y^2+xy^2}\right)$$ etc.
• This is more elegant than the answer I posted. After the first equality you immediately get $e^1$. – StarBug Jul 13 '19 at 9:27
Let $$\epsilon>0$$. Then for $$|x|<\epsilon$$: $$(1+x^2+y^2)^{\frac{1}{x^2+y^2+\epsilon y^2+\epsilon x^2}} \leq (1+x^2+y^2)^{\frac{1}{x^2+y^2+x y^2}} \leq (1+x^2+y^2)^{\frac{1}{x^2+y^2-\epsilon y^2-\epsilon y^x}}.$$ Since $$(1+x^2+y^2)^{\frac{1}{x^2+y^2-\epsilon y^2-\epsilon y^x}}= \Big((1+x^2+y^2)^{\frac{1}{x^2+y^2}}\Big)^{\frac{1}{(1-\epsilon)}} \rightarrow e^{\frac{1}{(1-\epsilon)}} \text{ as }(x,y)\rightarrow 0$$ and $$(1+x^2+y^2)^{\frac{1}{x^2+y^2+\epsilon y^2+\epsilon y^x}}= \Big((1+x^2+y^2)^{\frac{1}{x^2+y^2}}\Big)^{\frac{1}{(1+\epsilon)}} \rightarrow e^{\frac{1}{(1+\epsilon)}} \text{ as }(x,y)\rightarrow 0$$ it follows that $$e^{\frac{1}{(1+\epsilon)}} \leq \lim_{(x,y)\rightarrow 0} (1+x^2+y^2)^{\frac{1}{x^2+y^2+x y^2}} \leq e^{\frac{1}{(1-\epsilon)}}.$$ The above is true for all $$\epsilon>0$$. Therefore we must have $$\lim_{(x,y)\rightarrow 0} (1+x^2+y^2)^{\frac{1}{x^2+y^2+x y^2}} = e^1 = e.$$