Let $C_R$ denote the semicircle $z=Re^{i\theta}, 0\leq\theta\leq\pi$ from $z=R$ to $z=-R$, where $R>3$. Then show that $$\lim\limits_{R\to 0}\int_{C_R}\frac{(z+1)dz}{(z^2+4)(z^2+9)}=0$$
My attempt:
I tried using the triangle inequalities as follows:
$$|z+1|\leq|z|+|1|=R+1$$ $$|z^2+4|\geq|z^2|-|4|=R^2-4$$ $$|z^2+9|\geq|z^2|-|9|=R^2-9$$ $$\implies|f(z)|=\left|\frac{(z+1)}{(z^2+4)(z^2+9)}\right|\leq\frac{R+1}{(R^2-4)(R^2-9)}$$ Also, $|f(z)|\geq0$. So applying the ML inequality, we get
$$\int_{C_R}0\ dz\leq\int_{C_R}|f(z)|\ dz\leq\frac{R+1}{(R^2-4)(R^2-9)}\cdot\pi R$$
Applying $\lim\limits_{R\to0}$ to this inequality (attempt to use Sandwich theorem) yields the left side equal to $0$, but the right side ends up as a $\frac00$ form, which I'm unable to solve.
Im pretty sure there's a typo in your statement, because it's false. Also you have $R>3$ and you are asked about $R\to 0$? Does not, make any sense. Indeed, this integral goes to $\infty$ when $R\to 0$, so im sure that in fact you are asked to show what happens when $R\to\infty$, which is that the integral goes to zero, since your bounds are correct and $$\pi R\frac{R+1}{(R^2-4)(R^2-9)}\sim \frac{1}{R^2}\to 0,$$ as $R\to\infty$.