Linked Questions

93
votes
3answers
5k views

Are all limits solvable without L'Hôpital Rule or Series Expansion

Is it always possible to find the limit of a function without using L'Hôpital Rule or Series Expansion? For example, $$\lim_{x\to0}\frac{\tan x-x}{x^3}$$ $$\lim_{x\to0}\frac{\sin x-x}{x^3}$$ $$\...
10
votes
5answers
872 views

Determine $\lim_{x \to 0}{\frac{x-\sin{x}}{x^3}}=\frac{1}{6}$, without L'Hospital or Taylor

How can I prove that $$\lim_{x \to 0}{\frac{x-\sin{x}}{x^3}}=\frac{1}{6}$$ without using L'Hospital or Taylor series? thanks :)
7
votes
4answers
6k views

$\lim_{x\to0} \frac{x-\sin x}{x-\tan x}$ without using L'Hopital

$$\lim_{x\to0} \frac{x-\sin x}{x-\tan x}=?$$ I tried using $\lim_{x\to0} \frac{\sin x}{x}=1$. But it doesn't work :/
9
votes
3answers
6k views

Solving $\lim\limits_{x\to0} \frac{x - \sin(x)}{x^2}$ without L'Hospital's Rule.

How to solve $\lim\limits_{x\to 0} \frac{x - \sin(x)}{x^2}$ Without L'Hospital's Rule? you can use trigonometric identities and inequalities, but you can't use series or more advanced stuff.
18
votes
6answers
754 views

Is there a geometric method to show $\sin x \sim x- \frac{x^3}{6}$

I found a geometric method to show $$\text{when}\; x\to 0 \space , \space \cos x\sim 1-\frac{x^2}{2}$$ like below : Suppose $R_{circle}=1 \to \overline{AB} =2$ in $\Delta AMB$ we have $$\overline{AM}^...
3
votes
4answers
830 views

Calculate the limit : $\lim_{x \to 0}\frac{x-\sin{x}}{x^3}$ WITHOUT using L'Hopital's rule [duplicate]

I was given a task to find $$\lim_{x\to0}\frac{x-\sin{x}}{x^3}$$ at my school today. I thought it was an easy problem and started differentiating denominator and numerator to calculate the limit but ...
3
votes
4answers
223 views

$\lim_{x\to 0} \frac{\sin x - x}{x^2}$ without L'Hospital or Taylor

It is easy to see that $$\lim_{x\to 0} \frac{\sin x - x}{x^2} =0, $$but I can't figure out for the life of me how to argue without using L'Hospital or Taylor. Any ideas?
4
votes
3answers
167 views

Trying to find $\lim_{x \to 0} \frac{x - \sin x}{(x \sin x)^{(3/2)}}$ using L'Hopital's

I'm trying to use L'Hopital's rule to calculate: $$\lim_{x \to 0^+} \dfrac{x - \sin x}{(x \sin x)^{(3/2)}}$$ Taking a couple of derivatives of the denominator gets quite nasty, so I'd like to find a ...
6
votes
2answers
450 views

Solve $\lim_{x\to 0} \frac{\sin x-x}{x^3}$

I'm trying to solve this limit $$\lim_{x\to 0} \frac{\sin x-x}{x^3}$$ Solving using L'hopital rule, we have: $$\lim_{x\to 0} \frac{\sin x-x}{x^3}= \lim_{x\to 0} \frac{\cos x-1}{3x^2}=\lim_{x\to 0}...
7
votes
1answer
375 views

Computing: $L =\lim_{n\rightarrow\infty}\left(\frac{\frac{n}{1}+\frac{n-1}{2}+\cdots+\frac{1}{n}}{\ln(n!)} \right)^{{\frac{\ln(n!)}{n}}} $

Compute the following limit: $$L =\lim_{n\rightarrow\infty}\left(\frac{\frac{n}{1}+\frac{n-1}{2}+\cdots+\frac{1}{n}}{\ln(n!)} \right)^{{\frac{\ln(n!)}{n}}} $$ I'm looking for an easy, simple ...
2
votes
5answers
61 views

Limit $\lim_\limits{x\to0}{\tan{x}-\sin{x}\over x^3}$

What is the limit of: $$\lim_\limits{x\to0}{\tan{x}-\sin{x}\over x^3}$$ So what I tried is: $$\lim_\limits{x\to0}{{\sin{x}\over\cos{x}}-\sin{x}\over x^3}=\lim_\limits{x\to0}{{\sin{x}}({1\over\cos ...
1
vote
3answers
132 views

Find $\lim\limits_{x\to0}\frac{\sin(x)-\tan(x)}{\arcsin(x)-\arctan(x)}$ [duplicate]

Find $\displaystyle \lim_{x\to0}\frac{\sin(x)-\tan(x)}{\arcsin(x)-\arctan(x)}$ I tried using L'hopital's rule but it got very messy very fast UPDATE- So reading about the Taylor series this is what ...
1
vote
1answer
178 views

How to solve this limit question without using L'Hopital's rule?

Find the limit $$\lim_{x\rightarrow 0} \frac{\sin x-x}{\tan^3 x}$$ I found the limit which is $-\frac{1}{6}$ by using L'Hopital Rule. Is there another way to solve it without using the rule? Thanks ...
1
vote
2answers
199 views

limit of trigonometric functions without L'Hôpital's rule

$\lim_{x\to 0} [\csc^22x-\frac 1 {4x^2}]\ $ and $\lim_{x\to y} \frac {\tan x- \tan y} {1-\frac{x}{y}(\tan x\tan y+1)}\ $ We have not learn L'Hôpital's rule in my class, we do the $\lim_{x\to 0}\...
1
vote
1answer
178 views

How to solve $\lim\limits_{x\to 0} \frac{x - \tan(x)}{x^2}$ Without L'Hospital's Rule?

How to solve $\lim\limits_{x\to 0} \frac{x - \tan(x)}{x^2}$ Without L'Hospital's Rule? you can use trigonometric identities and inequalities, but you can't use series or more advanced stuff.

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