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I need help with this problem as stated in the title:

$$\lim_{x\to 0}\frac{\sin(\tan(x))-\sin(\sin(x))}{\tan(\tan(x))-\tan(\sin(x))}$$

I'm trying to convert the problem to using only the standard limits, with that said, I'm trying to solve the problem without using L'Hospital nor any Maclaurin series.

Do you guys have any tips on how to tackle the problem as I'm really struggling with this one.

Thank you.

K.defaoite
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Tanamas
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4 Answers4

3

Alternate approach:

\begin{align} \frac{\sin(\tan x)-\sin(\sin x)}{\tan(\tan x)-\tan(\sin x)} &= \frac{2\sin(\frac{\tan x-\sin x}{2})\cos(\frac{\tan x+\sin x}{2})}{\frac{\sin(\tan x-\sin x)}{\cos(\tan x)\cos(\sin x)}} \\ &= \frac{2\sin(\frac{\tan x-\sin x}{2})\cos(\frac{\tan x+\sin x}{2})}{2\sin(\frac{\tan x-\sin x}{2})\cos(\frac{\tan x-\sin x}{2})}\cdot \cos(\tan x)\cos(\sin x) \\ &= \frac{\cos(\frac{\tan x+\sin x}{2})}{\cos(\frac{\tan x-\sin x}{2})}\cdot \cos(\tan x)\cos(\sin x) \end{align}

and its limit as $x\to0$ can be evaluated easily.

RiverX15
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We can manipulate the given expression as follows

$$\frac{\sin(\tan(x))-\sin(\sin(x))}{\tan(\tan(x))-\tan(\sin(x))}=\frac{\frac1{\cos^3x }\frac{\sin(\tan(x))-\tan x}{\tan^3 x}-\frac{\sin(\sin(x))-\sin x}{\sin^3 x}+\frac{\tan x-\sin x}{\sin^3 x}}{\frac1{\cos^3x }\frac{\tan(\tan(x))-\tan x}{\tan^3 x}-\frac{\tan(\sin(x))-\sin x}{\sin^3 x}+\frac{\tan x-\sin x}{\sin^3 x}}$$

then we can use that as $x \to 0$

$$\frac{\sin x-x}{x^3}\to -\frac16$$

$$\frac{\tan x-x}{x^3}\to \frac13$$

with the methods indicated here:

user
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  • Thank you alot! But how in the world did you even come up with that manipulation of the expression. I see that's truly effective, but how? I'd never come up with such a thing, ever :) – Tanamas Sep 25 '21 at 14:31
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    @Tanamas As a first step you need to become confident with standard limits and then try to manipulate the expression to make emerge them. When the order involved is the first this always works well. Here we need to refer to higher order standard" limits to obtain the result but the way to proceed is the same. – user Sep 25 '21 at 14:35
  • Okay, thank you for the tip! Do you got any "resources" in terms of links where I can find more challenging limits like these? – Tanamas Sep 25 '21 at 14:39
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    @Tanamas I don't have specific resources to indicate other than MSE, you can find other similar limits here. – user Sep 25 '21 at 14:43
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This answer will use some trigonometric identities and the standard limits $$\lim_{y\to 0}\frac{\sin y}y =1 \text{ and } \lim_{y\to 0}\frac y{\tan y} =1\tag 1$$ Using $\sin C-\sin D=2\cos \frac{D+C}2\sin \frac{C-D}2$, numerator of the given expression becomes: $$2\sin \Big(\frac{\tan x-\sin x}{2}\Big)\cos \frac{\tan x+\sin x}{2}\tag 2$$ Using $\tan A- \tan B=(1+\tan A\tan B)\tan (A-B)$, denominator of the given expression becomes $$(1+\tan(\tan x)\tan(\sin x))\tan (\tan x-\sin x)\tag 3$$

So re-writing the given expression using $(1), (2)$ and $(3)$ gives: $$2\frac{\cos (\frac{\tan x+\sin x}{2})}{1+\tan(\tan x)\tan(\sin x)}.\frac{\sin \Big(\frac{\tan x-\sin x}{2}\Big)}{\frac{\tan x-\sin x}{2}}.\frac{\frac{\tan x-\sin x}{2}}{\tan x-\sin x}.\frac{\tan x-\sin x}{\tan (\tan x-\sin x)}\tag 4$$

By $(1)$, the first two terms in $(4)$ have limits (as $x\to 0$) $2$ and $1$ respectively and the last term has limit $1$ as $x\to 0$. The third term has limit equal to $\frac 12$ as $x\to 0$ so by limit rules the limit of the given expression is $2\times 1 \times \frac 12 \times 1=1$ as $x\to 0$.

Koro
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-1

Link to the answer image please check

The answer that i got for this problem is 1 without using L-Hospital or Maclurian series.

Saksham Chhimwal
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