So I had this question

$$\lim_{x \to 0} \frac{\sin(ax) + bx}{\sin(bx) + ax}$$

I did it this way:

$$\lim_{x \to 0} \frac{\sin(ax) + bx}{\sin(bx) + ax}$$

$$= \frac{\lim_{x \to 0}\sin(ax) + \lim_{x \to 0}bx}{\lim_{x \to 0}\sin(bx) + \lim_{x \to 0}ax}$$

$$= \frac{\lim_{x \to 0}\sin(ax)}{\lim_{x \to 0}\sin(bx)}$$

$$= \frac{\displaystyle\frac{\lim_{x \to 0}\sin(ax) \cdot ax}{ax}}{\displaystyle\frac{\lim_{x \to 0}\sin(bx) \cdot bx}{bx}}$$

$$=\frac{\lim_{x \to 0} ax}{\lim_{x \to 0} bx}$$

$$=\lim_{x \to 0}\frac{ax}{bx}$$

$$=\lim_{x \to 0}\frac{a}{b} = \frac{a}{b}$$

Which is incorrect.

I did it again and with a different approach :

$$\lim_{x \to 0} \frac{\sin(ax) + bx}{\sin(bx) + ax}$$

$$= \lim_{x \to 0} \frac{\displaystyle\frac{\sin(ax) + bx}{bx}}{\displaystyle\frac{\sin(bx) + ax}{bx}}$$

$$= \lim_{x \to 0} \frac{\displaystyle\frac{\sin(ax)}{bx} +1}{\displaystyle\frac{\sin(bx) + ax}{bx}}$$

$$= \frac{\lim_{x \to 0} \left(\displaystyle\frac{\sin(ax)}{bx} +1\right)}{\lim_{x \to 0} \left( \displaystyle\frac{\sin(bx) + ax}{bx} \right) }$$

$$= \frac{\lim_{x \to 0} \left(\displaystyle\frac{\sin(ax)}{bx}\right) +1}{\displaystyle\frac{a}{b} + 1}$$

$$= \frac{\lim_{x \to 0} \left(\displaystyle\frac{\sin(ax) \cdot ax}{bx \cdot ax}\right) +1}{\displaystyle\frac{a}{b} + 1}$$

$$= \frac{\displaystyle\frac{a}{b} +1}{\displaystyle\frac{a}{b} + 1} = 1$$

Which is the correct answer.

Can anyone tell me where was my first approach incorrect ?

I am guessing that this line

$$\frac{\lim_{x \to 0} ax}{\lim_{x \to 0} bx} = \lim_{x \to 0}\frac{ax}{bx}$$

is incorrect.

Thanks in advance.

  • $\begingroup$ When you distribuite the limit over the two addends you were not allowed to ignore the linear part $\endgroup$ – N74 Jun 5 '16 at 16:46
  • $\begingroup$ @N74 Is there a simple way of saying what you just said ? $\endgroup$ – A---B Jun 5 '16 at 16:50
  • $\begingroup$ Not from my smartphone... I fear. $\endgroup$ – N74 Jun 5 '16 at 16:53
  • 1
    $\begingroup$ You can separate the limit in parts only if they are convergent, i.e. if the limit exist for any part. But $\lim_{x\to 0} \frac{1}{\sin(ax)+bx}$ is divergent. I think that the second manipulation is wrong too. $\endgroup$ – Masacroso Jun 5 '16 at 17:12
  • $\begingroup$ @Masacroso Can you show how to do this without spiting limits ? because the solution given also involves spiting the limits on both denominator and numerator. $\endgroup$ – A---B Jun 5 '16 at 17:33

First of all, very well-asked question! That makes it easy for people to help out.

You are correct about the specific mistake you made in the first attempt. In detail, your first step is really the combination of two steps: $$ \lim_{x\to0} \frac{\sin(ax)+bx}{\sin(bx)+ax} = \frac{\lim_{x\to0} (\sin(ax)+bx)}{\lim_{x\to0} (\sin(bx)+ax)} = \frac{\lim_{x\to0} \sin(ax)+\lim_{x\to0} bx}{\lim_{x\to0} \sin(bx)+\lim_{x\to0} ax}. $$ In the first step, you're trying to use the following fact: if two functions have limits, and the bottom function's limit is not $0$, then the limit of their quotient is the quotient of the limits. (In other words: limits distribute over arithmetic, but only when the distributed version makes sense.) Here, $\lim_{x\to0} (\sin(bx)+ax)=0$, and so the first step is not valid. (Indeed, it results in a "0 over 0 indeterminate form".)

Edit: Yes, the second method is correct. You're essentially doing this: $$ \lim_{x\to0} \frac{\sin(ax)+bx}{\sin(bx)+ax} = \frac{\lim_{x\to0} (\frac{\sin(ax)}x+\frac{bx}x)}{\lim_{x\to0} (\frac{\sin(bx)}x+\frac{ax}x)} = \frac{\lim_{x\to0} \frac{\sin(ax)}x+\lim_{x\to0}\frac{bx}x}{\lim_{x\to0} \frac{\sin(bx)}x+\lim_{x\to0}\frac{ax}x}. $$ This is valid provided that each littler limit exists and provided that the denominator doesn't end up equalling $0$. The four limits in question are $a$, $b$, $b$, and $a$, so they all exist; as long as $a+b\ne0$, then this is valid. (As it happens, the case $b=-a$ is very simple.)

  • $\begingroup$ So is my second way correct ? it doesn't look so. $\endgroup$ – A---B Jun 5 '16 at 18:11

In short, you cannot naively "split" limits. A first reason for that is that they don't necessarily exist. For example, the sequence $0$ has a limit in infinity (here, 0) and $(-1)^n$ has no limit when $n\to\infty$, but $0=(-1)^n-(-1)^n$.

Also, both the numerator and denominator may have a limit, but what if the denominator's limit is 0? Then the ratio is not defined.

An efficient way to deal with such limits is to use Taylor's expansion. Here, $\sin(ax)+bx=ax-(ax)^3/6+bx+o(x^3)$ and $\sin(bx)+ax=bx-(bx)^3+ax+o(x^3)$, so, (assuming $a\neq -b$, otherwise, it's easy):

\begin{align}\dfrac{\sin(ax)+bx}{\sin(bx)+ax}&=\dfrac{ax+bx-a^3x^3/6}{bx+ax-b^3x^3/6}+o(x^2)=\dfrac{\frac{1}{b+a}(a+b-a^3x^2/6)}{\frac{1}{b+a}(b+a-b^3x^2/6)}+o(x^2) \\ &=1+\dfrac{b^3-a^3}{6(a+b)}x^2+o(x^2) & \\\end{align}

This proves that the limit in $0$ is $1$, and also that the way the fraction approaches 1 is parabolic is the neighbourhood of 0.


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