Evaluate the limit of (sin 3x + x)/(x + sin 5x) as x approaches 0.

Step-by-step solution

1. 1

   Identify the form. Both $\sin 3x+x$ and $x+\sin 5x$ vanish at $x=0$, so this is $\frac{0}{0}$.

2. 2

   Recall the key limit. $\lim_{x\to 0}\frac{\sin kx}{x}=k$ for any constant $k$, which follows from $\frac{\sin kx}{x}=k\cdot\frac{\sin kx}{kx}$ and $\lim_{u\to 0}\frac{\sin u}{u}=1$.

3. 3

   Divide top and bottom by $x$. For $x\neq 0$ the quotient equals $\frac{\frac{\sin 3x}{x}+1}{1+\frac{\sin 5x}{x}}$. Dividing by $x$ is the standard move because it converts every term into that known limit.

4. 4

   Take the limit term by term. The numerator tends to $3+1=4$ and the denominator to $1+5=6$; both limits exist and the denominator limit is nonzero, so the quotient rule for limits applies.

5. 5

   Compute the ratio. $\frac{4}{6}=\frac{2}{3}$.

6. 6

   Verify numerically. At $x=10^{-6}$ the original expression evaluates to $0.66666667$, matching $\frac{2}{3}$.