I have an equation:

$$ \displaystyle\lim_{ δx \rightarrow 0 } \dfrac{ \sin \left(\left( x+ \dfrac{ δx }{ 2 } \right) δx \right) }{ \left( x+ \dfrac{ δx }{ 2 } \right) δx } $$

The solution says that it is equal to $1$ because of the $ \dfrac{ \sin ( u ) }{ u } = 1 $ when ${ u }$ approaches $0$ limit theorem.

But I don't understand how that can be applied here, since the equation has both $δx$ and $x$, so how can the theorem still be applicable?


2 Answers 2


Let $t = \left( x+ \dfrac{ \delta x }{ 2 } \right) \delta x .$

Notice that:

$$\lim_{\delta x \to 0} t = 0.$$


$$\displaystyle\lim_{ \delta x \rightarrow 0 } \dfrac{ \sin \left(\left( x+ \dfrac{ \delta x }{ 2 } \right) \delta x \right) }{ \left( x+ \dfrac{ \delta x }{ 2 } \right) \delta x } = \lim_{t \to 0} \frac{\sin(t)}{t} = 1.$$

  • 4
    $\begingroup$ @Ennar I think in the question $\delta x$ is an "infinitesimal" and does not mean $\delta \cdot x$. $\endgroup$
    – Gary
    Commented Jun 27, 2023 at 10:39
  • 3
    $\begingroup$ @Ennar I wouldn't be terribly surprised if the original source used $\Delta x$ instead (which is a bit more standard) and OP just used the lower-case without recognizing the difference. $\endgroup$ Commented Jun 27, 2023 at 10:43
  • 3
    $\begingroup$ I don't understand how the original limit $limδx→0$ turned into $limt→0$ $\endgroup$ Commented Jun 27, 2023 at 12:12
  • 1
    $\begingroup$ @kronaemmanuel You are a new user here, so I think we should cut you some slack. Clicking the green check mark means you fully or at least mostly accept the answer to a question. Please only do this if you actually do understand and accept the answer. You can also uncheck the green check mark now if you want to (i.e. if you don't feel satisfied with this answer.) $\endgroup$ Commented Jun 27, 2023 at 21:01
  • 1
    $\begingroup$ What Adam said was what I mean to say. Thank you @AdamRubinson. $\endgroup$
    – David K
    Commented Jun 28, 2023 at 1:53

You are asking about the general limit

$$ \lim_{y\to L} \frac{\sin(u(y))}{u(y)} $$

where $u(y)\to 0$ as $y\to L\in\mathbb{R}$.

Well, so long as there does not exist a sequence $y_n$ in the domain of $u(y)$ such that $y_n \overset{n\to\infty}{\to} L$ and $u(y_n) = 0 $ for each $n\in\mathbb{N},$ then this is equivalent to there existing a neighbourhood around $L$ such that $u(y) \neq 0$ for all $y$ in this neighbourhood. Then in this case, I do not see why we would not have:

$$ \lim_{y\to L} \frac{\sin(u(y))}{u(y)} = 1. $$

I believe you could also replace $L$ with $+\infty$ or $-\infty$ from the start, and the above reasoning would also hold.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .