Help solve the following limit.

Question

I need help with the following limit:

$\displaystyle{\lim_{t \to 0} \cfrac{t-\textrm{sin}(2t)}{t-\textrm{sin}(3t)}}$

I need to solve it algebraically and not with the L'Hôpital rule.

Answer 1

We have that $\displaystyle{\lim_{t \to 0} \cfrac{t-\textrm{sin}(2t)}{t-\textrm{sin}(3t)}}=\displaystyle{\lim_{t \to 0} \cfrac{t}{t-\textrm{sin}(3t)}}-\displaystyle{\lim_{t \to 0} \cfrac{\textrm{sin}(2t)}{t-\textrm{sin}(3t)}}=L$.

Let $M=\displaystyle{\lim_{t \to 0} \cfrac{t}{t-\textrm{sin}(3t)}}$ and $N=\displaystyle{\lim_{t \to 0} \cfrac{\textrm{sin}(2t)}{t-\textrm{sin}(3t)}}$.

Then,

$M=\displaystyle{\lim_{t \to 0} \cfrac{t}{t-\textrm{sin}(3t)}}=\displaystyle{\lim_{t \to 0} \cfrac{t}{t\left(1-\cfrac{\textrm{sin}(3t)}{t}\right)}}=\displaystyle{\lim_{t \to 0} \cfrac{1}{1-\cfrac{\textrm{sin}(3t)}{t}}}=\cfrac{1}{1-3}=-\cfrac{1}{2}$

$N=\displaystyle{\lim_{t \to 0} \cfrac{\textrm{sin}(2t)}{t-\textrm{sin}(3t)}}=\displaystyle{\lim_{t \to 0} \cfrac{\textrm{sin}(2t)}{t\left(1-\cfrac{\textrm{sin}(3t)}{t}\right)}}=\displaystyle{\lim_{t \to 0} \cfrac{2}{1-\cfrac{\textrm{sin}(3t)}{t}}}=\cfrac{2}{1-3}=-1$

Now we have that $L=M-N=-\cfrac{1}{2}+1=\cfrac{1}{2}$.

Answer 2

Making the problem more general, consider $$y=\frac{t-\sin (a t)}{t-\sin (b t)}$$ and use Taylor series built at $t=0$ $$\sin(ct)=c t-\frac{c^3 t^3}{6}+O\left(t^5\right)$$ making $$y=\frac{(1-a) t+\frac{a^3 }{6}t^3+O\left(t^5\right) } {(1-b) t+\frac{b^3 }{6}t^3+O\left(t^5\right) }$$ Now, use the long division to get $$y=\frac{1-a}{1-b}+\frac{ a(a^2+b^3)-b(b^2+a^3)}{6(b-1)^3}t^2+O\left(t^4\right)$$ which shows the limit and also how it is approached.

For sure, this works only if $b \neq 1$.

Answer 3

Since this looks like a homework problem, I’ll give suggestions:

• Expand sin(2x), sin(3x) in terms of sin(x) and cos(x)
• Small angle approximation/ Taylor Series centered around c = 0
• Make use of the limit of $\frac{\sin(\theta)}{\theta}$ as $\theta$ approaches 0

Some of these may be easier/more helpful than others. I gave a variety that could be used for these types of problems.

Answer 4

$$\lim_{t\to 0}\frac {t-\sin 2t}{t-\sin 3t}=$$ $$=\lim_{t\to 0}\frac {1-2\frac {\sin 2t}{2t}}{1-3\frac {\sin 3t}{3t}}=$$ $$=\frac {1-2\cdot 1}{1-3\cdot 1}=$$ $$=\frac {1}{2}.$$