3
$\begingroup$

I need to evaluate $$\lim_{x \to 0}{\frac{\sin(4x)}{\sin(3x)}}$$ I solved this using L'Hospital's Theorem and I got 4/3
However, is there a way to do this without applying this theoerm?

Improve this question
$\endgroup$
  • 1
    $\begingroup$ Do you have available the limit $\lim_{y \to 0} \frac{\sin y}{y} = 1$? If so, you can simply factor the function in the given limit and make appropriate substitutions. $\endgroup$ – Travis Willse Feb 29 '16 at 15:04
2
$\begingroup$

As $$\lim_{x \to 0}\frac{\sin(x)}{x}=1$$ $$\lim_{x \to 0}{\frac{\sin(4x)}{\sin(3x)}}$$ can be written as $$ \frac{4}{3}\lim_{x \to 0}\frac{\sin(4x)}{4x}\frac{3x}{\sin(3x)} $$ $$=\frac{4}{3}$$

Improve this answer
$\endgroup$
6
$\begingroup$

Rewrite as $$ \frac{4}{3}\lim_{x \to 0}\frac{\sin(4x)}{4x}\frac{3x}{\sin(3x)} $$

Improve this answer
$\endgroup$
4
$\begingroup$

Since $$\sin t=t-\frac{t^3}{3!}+\frac{t^5}{5!}+\ldots=\sum_{k=0}^{\infty}\frac{(-1)^kt^{2k+1}}{(2k+1)!}$$ we have \begin{align} \frac{\sin(4x)}{\sin(3x)}&=\frac{4x-\frac{(4x)^3}{3!}+\frac{(4x)^5}{5!}+\ldots}{3x-\frac{(3x)^3}{3!}+\frac{(3x)^5}{5!}+\ldots}\\ &=\frac{4-\frac{4^3}{3!}x^2+\frac{4^5}{5!}x^4+\ldots}{3-\frac{3^3}{3!}x^2+\frac{3^5}{5!}x^4+\ldots}\\ \lim_{x\to 0}\frac{\sin(4x)}{\sin(3x)}&=\frac{4}{3} \end{align}

Improve this answer
$\endgroup$
3
$\begingroup$

By elementary trigonometry, with obvious shorthands:

$$\frac{\sin(4x)}{\sin(3x)}=\frac{\sin(2(2x))}{\sin(x+2x)}=\frac{2(2sc)(2c^2-1)}{s(2c^2-1)+c(2sc)}=\frac ss\frac{4c(2c^2-1)}{4c^2-1}\to\frac43.$$

More generally, one can use

$$L_{m+1}=\lim_{x\to0}\frac{\sin(x+mx)}{\sin(x)}=\lim_{x\to0}\frac{\sin(x)\cos(mx)}{\sin(x)}+\lim_{x\to0}\frac{\cos(x)\sin(mx)}{\sin(x)}=1+L_m.$$

Together with $L_0=0$, this gives $L_m=m$.

[It is easy to generalize to $m$ rational, by rescaling $x$, and even to $m$ real by squeezing $m$ between rationals.]

Of course,

$$\lim_{x\to0}\frac{\sin(px)}{\sin(qx)}=\lim_{x\to0}\frac{\sin(px)}{\sin(x)}\frac{\sin(x)}{\sin(qx)}=\lim_{x\to0}\frac{\sin(px)}{\sin(x)}\lim_{x\to0}\frac{\sin(x)}{\sin(qx)}=\frac pq.$$

Improve this answer
$\endgroup$
1
$\begingroup$

By Taylor's expansion around $0$, $x\mapsto\sin (tx)=tx+O(t^2)$, where $t \in \mathbb{R}$. Can you conclude?

Improve this answer
$\endgroup$
  • $\begingroup$ What is Taylor's expansion? $\endgroup$ – sai saandeep Feb 29 '16 at 15:19
  • $\begingroup$ Loosely, any analytic function $f$ may be written in the form $f(x)=\sum_{i=0}^{\infty} f^{(i)}(p) \frac {(x-p)^i}{i!}$, so the idea would be to consider the first terms of this series, as the other terms would eventually converge to zero. $\endgroup$ – gpr1 Feb 29 '16 at 15:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.