Question

Calculus 1 limits!! Can someone check my work???

1) lim (5x-3sinx0/(x)
x-->0

The answer options are: a) 0 b) 8 c) 2 d) does not exist

I got D but idk if that is right or if I did the right steps. Is that the same for anyone else?

2)lim (6x/2sinx)
x-->0
My teacher did an example similar to this and I did this: [sin(9x)/2x] = 9/2
Is that correct? It's only the second day of summer classes so I'm trying to get the hang of these homework assignments as they are a bit different from his examples in class!

Answer 1

\[\begin{align*}\lim_{x\to0}\frac{5x-3\sin x}{x}&=\lim_{x\to0}\frac{5x}{x}-\lim_{x\to0}\frac{3\sin x}{x}\\
&=5\lim_{x\to0}\frac{x}{x}-3\lim_{x\to0}\frac{\sin x}{x}\\
&=5\lim_{x\to0}1-3\lim_{x\to0}\frac{\sin x}{x}\\
&=5(1)-3(1)\\
&=2
\end{align*}\]
\[\begin{align*}\lim_{x\to0}\frac{6x}{2\sin x}&=3\lim_{x\to0}\frac{x}{\sin x}\\
&=3(1)\\
&=3
\end{align*}\]
Bonus:
\[\begin{align*}\lim_{x\to0}\frac{\sin9x}{2x}&=\frac{9}{2}\lim_{x\to0}\frac{\sin9x}{x}\cdot\frac{1}{9}\\
&=\frac{9}{2}\lim_{x\to0}\frac{\sin9x}{9x}\\
&=\frac{9}{2}
\end{align*}\]
The takeaway here is to remember that
\[\lim_{x\to0}\frac{\sin ax}{ax}=1\]
for \(a\not=0\). The strategy is to rewrite the given expression into a form that resembles this and to keep track of your coefficients and added terms.