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$$ \lim \limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1} $$

According to my textbook the limit equals $2$.

What I have tried:

Using the squeeze theorem:

$$ \lim \limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1}} \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1} \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 2} $$ $$ 0 \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1} \le 0 $$

I have also tried to use the squeeze theorem with two other equations and obtained different values:

$$ \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2 - 1}{\sqrt{x^2 +y^2 + 1}} \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1} \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2 + 1}{\sqrt{x^2 +y^2 + 1} - 2} $$ $$ -1 \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1} \le -1 $$

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  • $\begingroup$ Try multiplying the numerator and denominator by the conjugate of $\sqrt{x^2+y^2+1}-1$, which is $\sqrt{x^2+y^2+1}+1$. Algebraic manipulation like this can be helpful when you have something of the form $\sqrt a+b$ in the denominator of an expression. $\endgroup$
    – Jared
    Mar 29, 2013 at 0:06
  • $\begingroup$ Thank you, I can't believe I didn't think to do that. I've been struggling on that for awhile. $\endgroup$ Mar 29, 2013 at 0:10
  • $\begingroup$ You're very welcome. Glad to help. $\endgroup$
    – Jared
    Mar 29, 2013 at 0:15

3 Answers 3

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Remember the difference of squares algebraic identity.

$$ A^2 - B^2 = (A - B)(A + B) $$

Why is that useful? With $A = \sqrt{x^2 + y^2 + 1}$ and $B = 1$, the denominator of your expression is $A - B$. With that in mind,

$$ \begin{align} \frac{x^2 + y^2}{\sqrt{x^2 + y^2 + 1} - 1} &= \frac{x^2 + y^2}{\sqrt{x^2 + y^2 + 1} - 1} \cdot \frac{\sqrt{x^2 + y^2 + 1} + 1}{\sqrt{x^2 + y^2 + 1} + 1} \\ &= \frac{(x^2 + y^2)\left(\sqrt{x^2 + y^2 + 1} + 1 \right)}{x^2 + y^2} \\ &= \sqrt{x^2 + y^2 + 1} + 1 \end{align} $$

Now, you can evaluate the limit as $(x, y) \to (0, 0)$ simply by evaluation, since this expression is continuous at the origin.

$$ \lim_{(x, y) \to (0, 0)} \sqrt{x^2 + y^2 + 1} + 1 = \sqrt{0^2 + 0^2 + 1} + 1 = 2. $$

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    $\begingroup$ Thank you, that makes perfect sense. $\endgroup$ Mar 29, 2013 at 0:11
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The following "trick" can be useful. Let $x=r\cos\theta$, $y=r\sin\theta$. Then we are finding $$\lim_{r\to 0}\frac{r^2}{\sqrt{r^2+1}-1}.$$ This can be computed using any of the usual one variable techniques.

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  • $\begingroup$ Or even better, show that $\lim_{t\to 0} \frac{t}{\sqrt{t+1}-1} = 2$ and compose functions. $\endgroup$
    – wj32
    Mar 29, 2013 at 0:26
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    $\begingroup$ @wj32: Your suggestion is particularly nice because the limit, apart from being upside down, is a derivative. $\endgroup$ Mar 29, 2013 at 0:29
  • $\begingroup$ Hmm, I didn't notice that! $\endgroup$
    – wj32
    Mar 29, 2013 at 0:41
  • $\begingroup$ @AndréNicolas - what's the significance of it being a derivative? $\endgroup$ Mar 29, 2013 at 2:24
  • $\begingroup$ @VincentTjeng: No great significance, but many of the basic limits are connected with the derivative. So they can be identified immediately. If you take the reciprocal of this one, it is the derivative of $\sqrt{1+x}$ at $x=0$. $\endgroup$ Mar 29, 2013 at 5:09
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$$ \displaystyle \lim_{(x,y)\to(0,0)} \frac{(x^2+y^2)}{(x^2-y^2)}+ix $$

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