Limit of functions from $C^1$ Let $\gamma(t)=(\gamma_1(t),\gamma_2(t))\in C^1$ with $\gamma'(t_0)=(0,0)$ and $\gamma(t_0)=(0,0)$.
How to prove rigorously that $$\lim \limits_{t\to t_0}\dfrac{\gamma^4_1(t)\gamma_1'(t)+3\gamma^2_1(t)\gamma_2^2(t)\gamma_1'(t)-2\gamma^3_1(t)\gamma_2(t)\gamma_2'(t)}{[\gamma^2_1(t)+\gamma^2_2(t)]^2}=0.$$
Can anyone show the full proof please?
Here's my thought:
$$\left|\dfrac{\gamma_1^4(t)\gamma'_1(t)}{[\gamma^2_1(t)+\gamma^2_2(t)]^2}\right|\leqslant |\gamma'_1(t)|$$ but $|\gamma'_1(t)|\to 0$ as $t\to t_0$.
How to prove that the second part, i.e.$$\dfrac{3\gamma^2_1(t)\gamma_2^2(t)\gamma_1'(t)-2\gamma^3_1(t)\gamma_2(t)\gamma_2'(t)}{[\gamma^2_1(t)+\gamma^2_2(t)]^2}\to 0$$ as $t\to t_0$ ?

Use the following facts : 
$$\frac{ \gamma_1^4 }{|\gamma|^4} \leq 1
,\ 
\frac{ \gamma_1^2\gamma_2^2 }{|\gamma|^4} \leq 1
,\ 
\frac{ \gamma_1^3\gamma_2 }{|\gamma|^4} \leq 1$$
Note that there exist three terms in given limit So we will prove that each term has limit $0$ 
Firstly $$ \lim\ \bigg| \frac{ \gamma_1^4 }{|\gamma|^4} \gamma_1' \bigg|\leq \lim\ \bigg| \frac{ \gamma_1^4 }{|\gamma|^4} \bigg| |\gamma_1'| \leq \lim |\gamma_1'| =0 $$ 
Remaining two terms are proved by similar way
