find lim x→0 (tan x + 9x)/sin x.

Explanation:

Step1: Rewrite tan x

Recall that $\tan x=\frac{\sin x}{\cos x}$, so the limit becomes $\lim_{x\rightarrow0}\frac{\frac{\sin x}{\cos x}+9x}{\sin x}$.

Step2: Split the fraction

$\lim_{x\rightarrow0}\left(\frac{\frac{\sin x}{\cos x}}{\sin x}+\frac{9x}{\sin x}\right)=\lim_{x\rightarrow0}\left(\frac{1}{\cos x}+\frac{9x}{\sin x}\right)$.

Step3: Use limit properties

By the sum - rule of limits $\lim_{x\rightarrow a}(f(x)+g(x))=\lim_{x\rightarrow a}f(x)+\lim_{x\rightarrow a}g(x)$, we have $\lim_{x\rightarrow0}\frac{1}{\cos x}+\lim_{x\rightarrow0}\frac{9x}{\sin x}$.

Step4: Evaluate $\lim_{x\rightarrow0}\frac{1}{\cos x}$

Substitute $x = 0$ into $\frac{1}{\cos x}$, we get $\frac{1}{\cos(0)}=1$.

Step5: Evaluate $\lim_{x\rightarrow0}\frac{9x}{\sin x}$

Recall the well - known limit $\lim_{u\rightarrow0}\frac{\sin u}{u} = 1$, so $\lim_{x\rightarrow0}\frac{9x}{\sin x}=9\lim_{x\rightarrow0}\frac{x}{\sin x}=9\times1 = 9$.

Step6: Combine the results

$1 + 9=10$.

Answer:

$10$