Two Multivariable Limits help! Will someone please help me understand the following?


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*$\lim \limits_{(x,y) \to (0,0)} \dfrac{\tan y \cdot \sin^2(x-7y)}{x^2+y^2} $ which is the expression I get after doing the substitution $x-7=x , y-1=y $ to the expression:
$\lim \limits_{(x,y) \to (7,1)} \dfrac{\tan(y-1) \cdot \sin^2(x-7y)}{(x-7)^2+(y-1)^2} $. I tried dividing and and multiplying by the same expression to get the limit of $ \frac{y(x-7y)^2}{x^2+y^2} $ but I am not sure how to calculate this limit. 

*$\lim \limits_{(x,y) \to (0,0)} \dfrac{\sqrt{1+2xy}+7xy-1}{\sin(xy)} $ . After multiplying the entire expression by $\dfrac{ \sqrt{1+2xy} -7xy+1}{ \sqrt{1+2xy} -7xy+1 }  $ I get $\lim \limits_{(x,y) \to (0,0)} \dfrac{1+2xy-(7xy-1)^2}{\sin(xy) (\sqrt{1+2xy} -7xy+1) }$  and I don't think it gives me something.
Will you please help me?
Thanks a lot 
 A: We look at the second problem. There is a side issue of what happens when we approach along a path that has $x=0$ or $y=0$ infinitely often. Some people would say the limit does not exist. But let's sidestep that by assuming that $xy\ne 0$ on our path of approach.
The $\frac{7xy}{\sin(xy)}$ is no problem, it has limit $7$, by the standard fact that $\lim_{t\to 0}\frac{\sin t}{t}=1$.
So  we are only concerned with the limit of $\frac{\sqrt{1+2xy}-1}{\sin(xy)}$. Multiply top and bottom by $\sqrt{1+2xy}+1$, and everything turns out well.
Added: For the edited version of the first problem, make the change of variable $x-7s=s$, $y-1=t$. We end up with
$$\lim_{(s,t)\to (0,0)} \frac{\tan t \sin^2 s}{s^2+t^2}.$$
Multiply top and bottom by $s^2 t$. So we are interested in the behaviour of 
$$\frac{\tan t}{t}\frac{\sin^2 s}{s^2} \frac{s^2 t}{s^2+t^2}.$$
The fractions $\frac{\tan t}{t}$ and $\frac{\sin^2 s}{s^2}$ approach $1$. So we only care about
$$\lim_{(s,t)\to (0,0)}\frac{s^2 t}{s^2+t^2}.$$
This is easy. The standard thing is to let $s=r\cos\theta$ and $t=r\sin\theta$. The expression simplifies to $r\cos^2 \theta\sin\theta$, so the limit is $0$. 
