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Can someone help me calculate the following limits?

1) $ \displaystyle\lim _ {x \to 0 , y \to 0 } \frac{\sin(xy)}{\sqrt{x^2+y^2}} $ (it should equal zero, but I can't figure out how to compute it ) .

2) $\displaystyle\lim_ {(x,y)\to (0,\frac{\pi}{2} )} (1-\cos(x+y) ) ^{\tan(x+y)} $ (it should equal $1/e$).

3) $ \displaystyle\lim_{(x,y) \to (0,0) } \frac{x^2 y }{x^2 + y^4 } $ (which should equal zero).

4) $\displaystyle \lim_{(x,y) \to (0,1) } (1+3x^2 y )^ \frac{1}{x^2 (1+y) } $ (which should equal $e^{3/2}$ ).

Any help would be great !

Thanks in advance !

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Compute limit of variable x then variable y if it is the same as limit of y then x , you have a limit other wise it is meaningless limit. –  Arjang Jul 9 '12 at 6:12
    
What you're actually saying is that if I first take the limit on x and then on y, it should give me the same result as taking them both together to the limit? –  joshua Jul 9 '12 at 6:13
    
Well, it actually helps me in the first and third parts. But how can I know that the double limit exists in the first place? –  joshua Jul 9 '12 at 6:16
    
the double limit exists if the limit is not order dependent, that is the limit of x first then y is the same as limit of y first then x. –  Arjang Jul 9 '12 at 6:18
1  
@Arjang: are you sure about that? Take $\frac{xy}{x^2+y^2}$; if you compute the limit with $x$ first and then $y$ or with $y$ first and then $x$ it gives $0$ both times, but the limit doesn't exist (take the line $y=x$). –  Javier Badia Nov 1 '12 at 23:56

5 Answers 5

up vote 3 down vote accepted

Hints: For problem $1$, use the fact that $|\sin t|\le |t|$. Then to show that the limit of $\frac{xy}{\sqrt{x^2+y^2}}$ is $0$, switch to polar coordinates.

For problem $3$, it is handy to divide top and bottom by $x^2$, taking care separately of the difficulty when $x=0$.

For problem $2$, write $\tan(x+y)$ in terms of $\cos(x+y)$ and $\sin(x+y)$. Be careful about which side of $\pi/2$ the number $x+y$ is.

For problem $4$, it is useful to adjust the exponent so that it is $\frac{1}{3x^2y}$, or $\frac{1}{x^2y}$.

In $2$ and $4$, you may want to take the logarithm, and calculate the limit of that, though it is not necessary.

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Thanks a lot !!! Great hints ! –  joshua Jul 9 '12 at 6:56

Here I am using Andre’s idea to show that $\frac{xy}{\sqrt{x^2+y^2}}$ has limit $0$ when both $x$ and $y$ tend to $0$. For this we have to show: $$\forall \epsilon>0,∃ \delta>0, \forall (x,y), 0<||(x,y)-(0,0)||<\delta \longrightarrow |\frac{xy}{\sqrt{x^2+y^2}}-0|<\epsilon$$ Firstly, saying that $0<||(x,y)-(0,0)||<\delta$ is equivalent to $\sqrt{x^2+y^2}<\delta$ and therefore both of $|x|, |y|$ are less than $\delta$. If you take $z=max\{|x|,|y|\}$ then you have $z<\delta$ and : $$|\frac{xy}{\sqrt {x^2+y^2}}-0|=\frac{|x||y|}{ \sqrt{x^2+y^2}} ≤ \frac{zz}{ \sqrt{z^2+0}}=z<\delta $$ Now, it is enough to take $\delta$ as $\epsilon$. I hope mine could help you just for the first one. :)

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In this kinds of limits, you can take $r_\alpha(t)=(t,\alpha t)$ and put it into your function to find path wise limit of $f$ when $t$ tends to $0$. After simplifying the original function, if the limit of last expression approaches to zero then probably your original function has limit $0$ at $(0,0)$. Now, you have to use $\epsilon, \delta$ to prove your limit. Note that we see $Lim_{t\rightarrow 0} r_\alpha (t)=(0,0)$. –  Babak S. Jul 9 '12 at 7:09
    
+1 for you! :^) –  amWhy Mar 9 '13 at 2:57

For (4),

$$\lim_{(x,y)\to(0,1)}\left(1+3x^2y\right)^\frac{1}{x^2(1+y)}=\lim_{(x,y)\to(0,1)}\left(\left(1+3x^2y\right)^{\frac1{3x^2y}}\right)^{\frac{3y}{1+y}}\;,$$

where $$\lim_{(x,y)\to(0,1)}\left(1+3x^2y\right)^{\frac1{3x^2y}}$$ should be a familiar limit in the one-variable setting.

A similar trick works in (2) if you write the exponent $\tan(x+y)$ as $\dfrac1{\cos(x+y)}\cdot\sin(x+y)$.

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Thanks a lot !!! –  joshua Jul 9 '12 at 6:56

limit 1. $ \displaystyle\lim _ {x \to 0 , y \to 0 } \left| \frac{\sin(xy)}{\sqrt{x^2+y^2}} \right | \leq \displaystyle\lim _ {x \to 0 , y \to 0 } \frac{|xy|}{\sqrt{x^2+y^2}} $ now use inequality $a^2+b^2 \geq 2 |ab|$ $\displaystyle\lim _ {x \to 0 , y \to 0 } \frac{|xy|}{\sqrt{x^2+y^2}} \leq \frac{1}{2} \displaystyle\lim _ {x \to 0 , y \to 0 } \sqrt{x^2+y^2}$ which is going to $0$.

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For #2, If you let t = x+y then the limit becomes lim as t-->pi/2 in t alone. Now let L = the original limit and take log (ln=log, for me) of both sides using the log rule log (a^z) = a*log(z). Finally (I won't do the algebra for you) you end up with a limit, in t, which is -1. Keeping in mind that you applied the log, you must now reverse that step. I.E. if the original limit is L you now have log(L) = -1, so L = e^(-1) or 1/e.

This "log trick" is a very useful (and legitimate as long as you avoid log(0)) that brings the exponent sort of "down to earth". so you can handle things all on one level with much less anxiety! the same trick will work on #4)

(I am sorry I haven't learned how to make the text look "fancy" by using all the syntactical rules. I keep allowing the mathematics to take preference over the formatting. I would really appreciate some sort of hands on "tutorial" rather than a list of benign rules. I am a "learn by doing" kind of person.)

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Would you mind writing in TeX format? please? –  c.p. Jan 25 '13 at 16:17
    
I haven't learned it yet. I apologize./ I read an intro to LaTex last night and I'll be trying to use it soon. In the meantime, I can't resist answering questions. I will try next time. is there some sort of tutorial within stackexchange that would be helpful? –  DaveUM Jan 26 '13 at 13:42
    
It'S not that hard :). What you wrote here is mainly only missing a pair of $ enclosing the math. –  c.p. Jan 26 '13 at 23:52
    
@JorgeCampos Can you elaborate? –  DaveUM Jan 27 '13 at 8:29
    
Well, it's replacing, say, log(a^z) by $\log(a^z)$, etc. You can edit your own answer correcting that. Take a look at tobi.oetiker.ch/lshort/lshort.pdf –  c.p. Jan 27 '13 at 21:46

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