Evaluate: $\lim_{x\to 0} \dfrac {\sin 7x+ 5x}{7x +\sin 5x}$ Evaluate: $\lim_{x\to 0} \dfrac {\sin 7x+ 5x}{7x +\sin 5x}$

My Attempt:
$$=\lim_{x\to 0} \dfrac {\sin 7x +5x}{7x+\sin 5x}$$
It takes $\dfrac {0}{0}$ form when $x=0$.
$$=\lim_{x\to 0} \dfrac {\dfrac {\sin 7x+5x}{x}}{\dfrac {7x+\sin 5x}{x}}$$
$$=\lim_{x\to 0} \dfrac {\dfrac {\sin 7x}{7x} \times 7+5}{7+\dfrac {\sin 5x}{5x}\times 5}$$

How do I move on?
$$\lim_{\theta \rightarrow 0}\frac{\sin \theta}{\theta}=1$$
Therefore your answer is : $$\frac{7+5}{7+5}=1$$
Maclaurin series expansion of $$\sin\theta=\frac{\theta^1}{1!}-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\frac{\theta^7}{7!}...$$

Hence
$$\lim_{x\to 0} \dfrac {\dfrac {\sin 7x}{7x} \times 7+5}{7+\dfrac {\sin 5x}{5x}\times 5}$$
can be written as
$$\lim_{x\to 0} \dfrac {5+\dfrac {\sin 7x}{x}}{7+\dfrac {\sin 5x}{x}}$$
$$=\lim_{x\to 0} \dfrac {5+\dfrac {\frac{(7x)^1}{1!}-\frac{(7x)^3}{3!}+\frac{(7x)^5}{5!}-\frac{(7x)^7}{7!}...}{x}}{7+\dfrac {\frac{(5x)^1}{1!}-\frac{(5x)^3}{3!}+\frac{(5x)^5}{5!}-\frac{(5x)^7}{7!}...}{x}}$$
$$=\lim_{x\to 0} \dfrac {5+7-\frac{7^3x^2}{3!}+\frac{7^5x^4}{5!}-\frac{7^7x^6}{7!}...}{7+5-\frac{5^3x^2}{3!}+\frac{5^5x^4}{5!}-\frac{5^7x^6}{5!}...}$$
$$=\dfrac {5+7}{7+5}$$
$$=\dfrac {12}{12}$$
$$=1$$