# How to prove $\,\lim_{(x,y)\to(0,0)}(1+x^2+y^2)^{1/(x^2+y^2+xy^2)}\,$ exists [closed]

My question is a straightforward one. How do I prove the following limit exists? $$\lim_{(x,y)\to(0,0)}(1+x^2+y^2)^{1/(x^2+y^2+xy^2)}$$

I imagine I ought to use the squeeze theorem in some way, but I cannot figure out how! I would be very thankful for your help.

• Also, I imagine it would be sufficient to set $x = t^{2}$ and $y=0$, but this piqued my interest.
– Carl
Jan 26 at 11:29
• Please do not use pictures for critical parts of your post as they may not be legible to some. If you need help in formatting the maths, here is a tutorial. Jan 26 at 11:49
• Does this answer your question? Limit of $(1+x^2+y^2)^{\frac{1}{x^2+y^2+x y^2}}$, where do I get it wrong? Jan 28 at 15:40

We have

$$\displaystyle{\left(1+x^2+y^2\right)^{\dfrac{1}{x^2+y^2+xy^2}}={\large{e}}^{\displaystyle{\left(\dfrac{\ln{ \left(1+x^2+y^2\right)}}{x^2+y^2+xy^2}\right)}}}.$$

We define $$f(x,y):= \frac{1}{x^2+y^2+xy^2} \cdot \ln{ \left(1+x^2+y^2\right)}.$$

We proceed with polar coordinates:

$$f\left(r \cos t, r \sin t\right)= \frac{1}{1+r \cos t \sin^2 t}\cdot \frac{\ln \left(1+r^2\right)}{r^2}.$$ Now we have

$$\displaystyle{\lim\limits_{r \to 0}\left(\frac{1}{1+r \cos t \sin^2 t}\right) \longrightarrow 1} \\ \qquad \\ \qquad \\ \text{and} \\ \qquad \\ \qquad \\ \lim\limits_{r \to 0}{\left(\frac{\ln \left(1+r^2 \right)}{r^2}\right)} \longrightarrow 1$$

Hence,

$$\lim\limits_{r \to 0}{f\left(r \cos t, r \sin t\right)} \longrightarrow 1 \\$$

$$\displaystyle{\therefore \lim\limits_{(x, y) \to (0, 0)}{\left(1+x^2+y^2\right)^{\dfrac{1}{x^2+y^2+xy^2}}} \longrightarrow e}.$$

• It was a (low-quality question and) duplicate. Jan 28 at 15:42