limit question

[rich stirrup]

hello how could i solve this question without using taylors expansion and lhopital rule i am in dire need so any help is appreciated the answer should be 0

[rich stirrup]

also is tan(x)/x exactly 1 or not?

as x approaches to 0

[strange lantern]

Take limit it's equal to 1. Sometimes we take tan(x)=x when x<10 degrees in physics class

[rich stirrup]

would this be an incorrect solution since tan(x)/x doesnt exactly equal 1 algebraically

[chrome atlas]

tan(x)/x should approach 1 when x -> 0 tho

it’s correct

tan(x) is basically x near x=0
same slope and all that

[rich stirrup]

i have asked chatgpt and deepseek they claim its incorrect and i remember something like that from class as well

like as x approaches 0 the limit gets closer to 1 but it never exactly reaches it

[chrome atlas]

desmos looks like the limit is 0

[rich stirrup]

it is 0 and is solvable with taylors and lhopital but our instructor asked us not to use them

[solar flame]

It is incorrect

[craggy mortar]

try l'hopital's rule

[strange lantern]

He can't

[rich stirrup]

we are not allowed to use l'hopital or taylor series

[craggy mortar]

Oh well squeeze lemma it is then

[rich stirrup]

instructor told us that there could be something to do with tan2x definition

[craggy mortar]

probably a substitution of some sort.

$$\tan(2x)=\frac{2\tan(x)}{1+\tan^2(x)}$$

[sturdy fjord]

Calistωo

[craggy mortar]

actually no

it's a bound

that's what it is

take the interval from $\left(-\frac{\pi}{4},\frac{\pi}{4}\right)$

[sturdy fjord]

Calistωo

[craggy mortar]

$$-\tan(2x)\leq \tan(x) \leq\tan(2x)$$

[sturdy fjord]

Calistωo

[craggy mortar]

$$\frac{-\tan(2x)-x}{ x^2} \leq \frac{\tan(x)-x}{x^2} \leq \frac{\tan(2x)-x}{ x^2}$$

[sturdy fjord]

Calistωo

[craggy mortar]

$$x^2 \leq 1+\tan^2(x)$$
$$\frac{\tan(x)-x}{1+\tan^2(x)} \leq \frac{\tan(x)-x}{x^2}$$

[sturdy fjord]

Calistωo

[craggy mortar]

$$\frac{\tan(x)-x}{1+\tan^2(x)} \leq \frac{\tan(x)-x}{x^2}\leq \tan^2(x)$$

[sturdy fjord]

Calistωo

[solar flame]

I don't know how exectly you've done limits because I can't exectly see how you are supposed to solve this via tan(2x) but I'll present my solution via standard limits results like
Those hold when x goes to 0
Limit of sinx/x=1
Limit of (1-cosx)/x^2 =1/2 or limit of (1-cosx)/x=0
Limit of (sinx-x)/x^3=-1/6
So (tanx-x)/x^2 is (sinx-xcosx)/(x^2cosx)
Add and subtract x
Sinx-x-xcosx+x=(sinx-x)+x(1-cosx) so you get
(tanx-x)/x^2=(sinx-x)/(x^2cosx) +(1-cosx)/(xcosx)
Let's take a look at the first limit
Multiply and divide by x you get
(sinx-x)/x^3 * x/cosx = (-1/6)0=0
Do the same for the second one
(1-cosx)/x^2 * x/cosx=1/20

Result is then 0

[craggy mortar]

that works too

probably a lot simplier than trying to find bounds

[solar flame]

If this inequality holds then yea from here the same results follows via squeeze lemma

[solar flame]

True

[craggy mortar]

I know that the lower bound is correct

the upper bound is definitely eh

I think if you pick $$\sin^2(x)$$

because in the interval $$\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$$ $$\tan(x)-x \leq \cos(x)$$

[sturdy fjord]

Calistωo

[craggy mortar]

$$x^2 \leq 1+\tan^2(x)$$
$$\frac{\tan(x)-x}{1+\tan^2(x)} \leq \frac{\tan(x)-x}{x^2} \leq \frac{\cos^2(x)}{x^2}$$

wait that goes to 1 :(

[sturdy fjord]

Calistωo

Calistωo

[solar flame]

Cosx/x would be 1/0 tho

[hybrid hemlock]

All you have to do is prove this:

$\newline 0\le\frac{\tan\left(x\right)-x}{x^{2}}\le x\newline$

for small positive x and then use the squeeze theorem.

I'll get you started:

$\newline 0\le\tan\left(x\right)\le x^{3}+x\newline \dots$

[sturdy fjord]

solarunes

[rich stirrup]

dont i need taylor expansion to prove tan(x) <= x+x^3

[hybrid hemlock]

No. Differentiation is enough.

Consider x³ + x - tan(x). This is equal to 0 at x=0. Now you just need to verify that it's not decreasing for small x.

[rich stirrup]

okay thanks for helping i will think about it after a rest also we can only use differentiation through limit definition

i also want to ask if this solution from chatgpt makes sense or not for anyone interested

actually

[hybrid hemlock]

It doesn't justify why tan(x) > x, which is what you'll need to prove for the inequality to work.

[rich stirrup]

i just got a solution from one of our instructors

it starts from page 2

[hybrid hemlock]

That works I guess. Not exactly the cleanest solution but a solution nonetheless.

[rich stirrup]

yep also thanks to everyone here helping to solve

[hybrid hemlock]

Type .solved to mark this as solved if you don't need help anymore.

[rich stirrup]

.solved