# Are there more solutions than this? Limit Problem

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
The answer is correct, but I have a couple of remarks. First, you wrote two times that $\sin(x)/x$ literally equals 1. In fact, $\sin(x)/x\ne1$ for all $x\in\mathbb{R}$. The familiar relationship between $\sin(x)/x$ and 1 always involves limits.

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 $(x_0,y_0)$.

#### goku900

##### New member
The answer is correct, but I have a couple of remarks. First, you wrote two times that $\sin(x)/x$ literally equals 1. In fact, $\sin(x)/x\ne1$ for all $x\in\mathbb{R}$. The familiar relationship between $\sin(x)/x$ and 1 always involves limits.

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 =/

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
If you want to justify $\lim_{x\to0}\sin(x)/x$ using Taylor series, then you say that by Taylor's theorem,
$\sin x=x+\alpha(x)x$
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,
$\lim_{(x,y)\to(0,1)}\frac{y\sin x}{x(y+1)} = 1\cdot\frac{1}{2}=\frac{1}{2}$

#### goku900

##### New member
If you want to justify $\lim_{x\to0}\sin(x)/x$ using Taylor series, then you say that by Taylor's theorem,
$\sin x=x+\alpha(x)x$
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.