1
$\begingroup$

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 $r→0$?

Thanks in advance!

$\endgroup$
  • $\begingroup$ Is this $$(1+x^2+y^2)^{\frac{1}{x^2+y^2+xy^2}}$$? $\endgroup$ – Dr. Sonnhard Graubner Jul 13 '19 at 8:50
  • $\begingroup$ Yes, i'm so sorry for the poor formatting. $\endgroup$ – Fosorf Jul 13 '19 at 8:51
  • $\begingroup$ Also, exact duplicate to Limit as (x,y) approaches (0,0) of ... $\endgroup$ – Lutz Lehmann Jul 13 '19 at 9:13
  • $\begingroup$ 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 $e$. $\endgroup$ – Imago Jul 13 '19 at 9:24
  • 1
    $\begingroup$ MMA says the result is given by $$\sqrt{e}$$ $\endgroup$ – Dr. Sonnhard Graubner Jul 13 '19 at 10:07
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ This is more elegant than the answer I posted. After the first equality you immediately get $e^1$. $\endgroup$ – StarBug Jul 13 '19 at 9:27
0
$\begingroup$

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 $$ \lim_{(x,y)\rightarrow 0} (1+x^2+y^2)^{\frac{1}{x^2+y^2+x y^2}} = e^1 = e.$$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.