$$\lim_{x \to 2} \frac{\cos{\left(\frac{\pi}{x}\right)}}{x-2}$$  This limit is supposed be found without L'Hospital's Rule, but I have not been able to get close to the answer using conjugates, squares, Pythagorean Identity, half angle formulas or Squeeze Theorem. For each attempt, I expected at least one of  the well-known limits $\lim_{x \to 0} \frac{\sin{x}}{x}$ or $\lim_{x \to 0} \frac{\cos{x} - 1}{x}$ to come into use. They frequently did in my attempts but did not go anywhere. I have also been experimented substitutions but in vain.  How can this be solved without using L'Hospital's Rule?
122631
I am a kid trying to teach myself Calculus in order to prepare for next year.   I have the expression  $$\lim_{x\to 2}\frac{\cos(\frac \pi x)}{x-2} $$  There is a hint that says to substitute t for $(\frac \pi2 - \frac \pi x)$ and WolframAlpha evaluates this expression as $\frac \pi4$. However, I got the answer of $1$. Can someone clarify the steps to solving this problem.
38715
Evaluate $$\lim_{x\to \infty} \cos (\sqrt {x+1})-\cos (\sqrt x)$$   For this question I have tried using the squeeze theorem and some trigonometric manipulations such as using $$2\sin A\sin B=\cos (A-B) -\cos (A+B)$$.  But they were of no use. I also tried using the substitutions like $$x=1/y$$ Hence as $x\to \infty$ , $y\to 0$ but still no success.  I could've also tried using L'Hospital rule but I am not able to convert the function in appropriate indeterminate form. Any help would be greatly appreciated
1999
This limit seemed quite unusual to me as there aren't any intermediate forms or series expansions which are generally used in limits. Stuck on this for a while now .Here's how it goes :    $$ \lim \limits_{n \to \infty} \left[\cos\left(x \over 2\right)\cos\left(x \over 4\right)       \cos\left(x \over 8\right)\ \cdots\ \cos\left(x \over 2^{n}\right)\right] $$
109450
$$\lim_{x\to y}\frac{x^y-y^x}{x^x-y^y} $$ I've tried L'Hospital and  Both doesn't tends to work. Please help someone.
34157
I'm trying to solve this limit     $$\lim_{x\to 0} \frac{\sin x-x}{x^3}$$   Solving using L'hopital rule, we have:  $$\lim_{x\to 0} \frac{\sin x-x}{x^3}= \lim_{x\to 0} \frac{\cos x-1}{3x^2}=\lim_{x\to 0} \frac{-\sin x}{6x}=\lim_{x\to 0} \frac{-\cos x}{6}=-\frac{1}{6}.$$  Am I right?  I'm trying to solve this using change of variables, I need help.  Thanks  EDIT  I didn't understand the answer and the commentaries, I'm looking for an answer using change of variables.
106621
I am trying to calculate the following limit without L'Hôpital's rule:     $$\lim_{n \to \infty} \dfrac{n}{\ln(n)}$$   I tried every trick I know but nothing works. You don't have to prove it by definition.
73476
Find: $$\lim\limits_{x \to 1} \left( {\frac{x}{{x - 1}} - \frac{1}{{\ln x}}} \right) $$  Without using L'Hospital or Taylor approximations  Thanks in advance
39067
Is there a way of showing this limit without using L'Hopital's rule? Specifically using the substitution $u = -\log{x},$ with knowledge of the limit of $\lim_{x\to0+}\log{x}.$ Thank you.
94214
I was solving limits and it said $$\lim_{x\to 0} \left[\frac{x^2}{\tan(x)\sin(x)}\right]=0$$  I tried thinking about it but since I am not really good at math I couldn't proceed.  I know that $\lim_{x\to 0} \left[\frac{\sin(x)}{x}\right] = 0$ and $\lim_{x\to 0} \left[\frac{\tan(x)}{x}\right] = 1$
86721
Calculate $$\lim_{x\to 0}\dfrac{1}{x^{\sin(x)}}$$  I'm pretty much clueless here, only that there is L'hospital obviously here.   Would appreciate any help.
101292
$$\lim_{x \to \frac{2}{\pi}}\lfloor \sin \frac{1}{x} \rfloor=0$$    I need to prove the limit of this using the $\epsilon - \delta $ way but I don't know how to find $\delta$ when I'm given a trigonometric function I know only how to do it with polynomial functions
45937
$$\lim_{x\to\infty}\left(\sin{\frac 1x}+\cos{\frac 1x}\right)^x$$  It is about the $(\to1)^{(\to\infty)}$ situation. Can we find its limit using the formula $\lim_{x\to\infty}(1+\frac 1x)^x=e$? If yes, then how?
108666
Prove that   $$\lim_{x\to \frac{\pi}{2}} \frac{1}{\big(x-\frac{\pi}{2}\big)}+{\tan(x)}=0.$$   I'm not really sure how to proceed. I know that I should not try L'Hôpital's rule (tried that) but not sure how I would incorporate into the Squeeze Theorem or how I would use continuity.  Thanks!  Edit: Turns out I was really dumb and you do use L'Hôpital's rule twice. I made the mistake of differentiating the whole quotient rather than the function on top and the bottom of the vinculum separately.
27207
$$\lim _{x \rightarrow 0} \left(\frac{ \sin x}{x}\right)^{1/x}$$  I have spent an hour on the above limit and have no work to show. I tried using L'Hopital's Rule, but just kept going around in circles. Any help would be appreciated. Thank you.
115213
$$\lim_{x\to 0}\frac{\sqrt[5]{1+\sin(x)}-1}{\ln(1+\tan(x))}$$  How to evaluate the limit of this function without using L'Hôpital's rule?
69608
I know with l'Hopital's rule it becomes $-\sin(x)$ which has the limit $0$. However, I have been wondering how to evaluate this limit without l'Hopital's rule.
55505
Consider the Limit  $\frac{x+\sin x}{x-\sin x}$  as $x  \rightarrow \infty$. Clearly this limit exists and is equal to $1$. Also notice that the function is of the form  $\frac{\infty}{\infty}$ and thus, L'Hopitals rule can be applied here. But on applying L'Hopitals rule, I am getting  Limit  $\frac{1+\cos x}{1-\cos x}$ as $x  \rightarrow \infty$, which is obviously non-converging. Where did I go wrong?