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- Thread starter goku900
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- Jan 30, 2012

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Second, in the last line you seem you use the fact that \[

\lim_{(x,y)\to(x_0,y_0)} f(x,y)=\lim_{y\to y_0} f\left(\lim_{x\to x_0} f(x,y), y\right)

\]

i.e., that you can take limits consecutively. I believe this is not even true in general when $f$ is continuous at $(x_0,y_0)$.

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- #3

Second, in the last line you seem you use the fact that \[

\lim_{(x,y)\to(x_0,y_0)} f(x,y)=\lim_{y\to y_0} f\left(\lim_{x\to x_0} f(x,y), y\right)

\]

i.e., that you can take limits consecutively. I believe this is not even true in general when $f$ is continuous at $(x_0,y_0)$.

So, how would I rewrite this? I'm not really sure =/

- Jan 30, 2012

- 2,571

\[

\sin x=x+\alpha(x)x

\]

where $\alpha(x)\to0$ as $x\to0$. Then $\sin(x)/x=1+\alpha(x)$, so

\[

\lim_{x\to0}\sin(x)/x= \lim_{x\to0}(1+\alpha(x)) =\lim_{x\to0}1+ \lim_{x\to0}\alpha(x)=1

\]

Using small-o notation,

\[

\lim_{x\to0}\sin(x)/x= \lim_{x\to0}(x+o(x))/x

=\lim_{x\to0}(1+o(1))=1

\]

Concerning the last line, I just wasn't sure what fact you relied on. You can split the limit into the product of limits because both of them exist.

\[

\lim_{(x,y)\to(0,1)}\frac{y\sin x}{x(y+1)} =

\left(\lim_{(x,y)\to(0,1)}\frac{\sin x}{x}\right) \left(\lim_{(x,y)\to(0,1)}\frac{y}{y+1}\right)

\]

Now since both functions depend on only one argument, we can use

\[

\lim_{(x,y)\to(x_0,y_0)}f(x) =\lim_{x\to x_0}f(x)

\]

and similarly for $y$, which follow easily from the definition of limits. So,

\[

\lim_{(x,y)\to(0,1)}\frac{\sin x}{x}= \lim_{x\to0}\frac{\sin x}{x}=1

\]

and

\[

\lim_{(x,y)\to(0,1)}\frac{y}{y+1}= \lim_{x\to1}\frac{y}{y+1}=\frac{1}{2}

\]

All in all,

\[

\lim_{(x,y)\to(0,1)}\frac{y\sin x}{x(y+1)} =

1\cdot\frac{1}{2}=\frac{1}{2}

\]

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- #5

\[

\sin x=x+\alpha(x)x

\]

where $\alpha(x)\to0$ as $x\to0$. Then $\sin(x)/x=1+\alpha(x)$, so

\[

\lim_{x\to0}\sin(x)/x= \lim_{x\to0}(1+\alpha(x)) =\lim_{x\to0}1+ \lim_{x\to0}\alpha(x)=1

\]

Using small-o notation,

\[

\lim_{x\to0}\sin(x)/x= \lim_{x\to0}(x+o(x))/x

=\lim_{x\to0}(1+o(1))=1

\]

Concerning the last line, I just wasn't sure what fact you relied on. You can split the limit into the product of limits because both of them exist.

\[

\lim_{(x,y)\to(0,1)}\frac{y\sin x}{x(y+1)} =

\left(\lim_{(x,y)\to(0,1)}\frac{\sin x}{x}\right) \left(\lim_{(x,y)\to(0,1)}\frac{y}{y+1}\right)

\]

Now since both functions depend on only one argument, we can use

\[

\lim_{(x,y)\to(x_0,y_0)}f(x) =\lim_{x\to x_0}f(x)

\]

and similarly for $y$, which follow easily from the definition of limits. So,

\[

\lim_{(x,y)\to(0,1)}\frac{\sin x}{x}= \lim_{x\to0}\frac{\sin x}{x}=1

\]

and

\[

\lim_{(x,y)\to(0,1)}\frac{y}{y+1}= \lim_{x\to1}\frac{y}{y+1}=\frac{1}{2}

\]

All in all,

\[

\lim_{(x,y)\to(0,1)}\frac{y\sin x}{x(y+1)} =

1\cdot\frac{1}{2}=\frac{1}{2}

\]

That makes perfect sense ! Awesome thank you so much.