I am trying to figure out how to make $\lim_{x \to 0} (1-2x)^{1/x}$ into $\frac{\infty}{\infty}$ or $\frac{0}{0}$ form for L'Hospital's Rule.   Using Desmos.com I have found that $(1-2x)^{1/x} \neq 1-2^{1/x}x^{1/x}$ and I'm not sure why that is either. That is the only thing I can think of to change the function's form.
123600
I have recently come across a problem concerning the limit of a sequence given as, $$\lim_{n\rightarrow \infty} (a^n + b^n)^\frac{1}{n}$$  Where $0&lt;a&lt;b$. Solving the limit is not a concern. But applying L'Hospital's rule to this limit is one to me.   On rearranging we get, $b \lim_{n\rightarrow \infty} ((\frac{a}{b}) ^n + 1)^\frac{1}{n}$. The solver had then applied the L'Hospital's rule to evaluate the limit. Here, as $n\rightarrow \infty$, the limit approaches the form $1^0$. I have tried to think about this limit approaching an indeterminate form. It does not. From my knowledge of Cauchy's Mean Value theorem, I say that we cannot apply the rule to determinate forms. But it seems to me that we can because it worked for the solver. I think I am not clear on this. Can somebody help?
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?
71898
As the title says, I want to show that the limit of  $$\lim_{x\to 0} \frac{\cos(x)}{x}$$  doesn't exist.  Now for that I'd like to show in a formally correct way that   $$\lim_{x\to 0^+} \frac{\cos(x)}{x} = +\infty$$  I'm sure this is right since $\displaystyle\lim_{x\to 0^+} \cos(x) = 1$ and $\displaystyle\lim_{x\to 0^+} x = 0$, but since $\displaystyle\lim_{x\to 0^+} x = 0$ I can't just say:  $$\lim_{x\to 0^+} \frac{\cos(x)}{x} = \frac{1}{\displaystyle\lim_{x\to 0^+} x}$$   Things I tried: expand the fraction and use L'Hospital's rule (This didn't seem to yield results even though I tried quite a few things, should it? If it does, should it always work? It worked for similar problems) and going by the definition of the limit which didn't work for me either.  I'm grateful for any hints for doing this problem and similar problems especially.
70158
Find $$\lim_{x \to 0} (\frac{\tan(x)}{x})^{\frac{1}{x}}$$   My first idea to solve this was to try to evaluate it and then apply the L'Hospital's rule. This is what I managed to achieve: $$\left(\frac{\tan(x)}{x}\right)^\frac{1}{x}=e^{\frac{1}{x}\ln\left(\frac {\tan(x)}x \right)} $$ However, this is problematic, because the L'Hospital's rule is not applicable in the exponent, therefore my transformation was not a good shot. What transformation should I apply to get the desired form?
104076
How to evaluate the following limit $$\displaystyle\lim_{x\to 0} \dfrac{e-\left(1 + x\right)^{1/x}}{\tan x}$$ I have tried to solve it using L-Hospital's Rule, but it creates utter mess. Thanks for your generous help in advance.
110601
$$\lim\limits_{x\to 1}\frac{\sqrt x-x^2 }{1-\sqrt x}$$  I'm not entirely sure how to go about this, since rationalizing anything doesn't quite work for me. Just looking for some help.
95216
So I ran into this problem:  $\lim _{x\to \pi }(\frac{\cos\left(\frac{x}{2}\right)}{\pi ^2-x^2})$  We haven't studied L'Hôpital's rule in my class yet, we basically need to transform the expression using trigonometric identities and use known limits like $\lim_{x \to 0} \frac{sinx}{x} = 1$ to calculate in case we get an indefinite form.  I couldn't find anything I could do to calculate this limit :(  Some help?  Thanks.
29562
I tried l'Hospital but that will require a lot (and I mean A LOT!!!) of differentiating  Is there a shortcut? $$\mathop {\lim }\limits_{x \to 0} \left( {{1 \over {{{\sin }^2}x}} - {1 \over {{x^2}}}} \right)$$  Thanks in advance
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.
110724
I am asked to calculate the limits of $\dfrac{x}{\sqrt{1+x^2}}$ for $x\to\infty$ and $x\to-\infty$  When I simplify the radical for $x\to\infty$, I generate:  $\dfrac{x}{\sqrt{1+x^2}} = \sqrt{\dfrac{x^2}{1+x^2}}$, after which I can use l'hopital the two parts because they are $\infty / \infty$.   I do lim lim $x\to-\infty$ in similar fashion, but here an extra minus appears in the simplying radicals step that I can't quite place. I never actually use that $x\to -\infty$, so I'm kind of lost how that gets there?
49802
I used $$(n!)^{\frac{1}{n}}=e^{\frac{1}{n}\ln(n!)}=e^{\lim\limits_{n\to\infty}\frac{1}{n}\ln(n!)}$$ Then using Stirling's approximation and L'Hospital's rule on $$\lim\limits_{n\to\infty}\frac{\ln(n!)}{n}$$ I get $$\lim\limits_{n\to\infty}\frac{\ln(n!)}{n}=\lim\limits_{n\to\infty}(\ln(n)+\frac{n+\frac{1}{2}}{n}-1)=\infty$$ Now, $$e^{\lim\limits_{n\to\infty}\frac{1}{n}\ln(n!)}=e^{\infty}=\infty$$ Thus $$\lim\limits_{n\to\infty}\frac{1}{\sqrt[n]n}=\frac{1}{\infty}=0$$  Is this correct approach and what other methods could be used?
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.
35223
I am able to evaluate the limit $$\lim_{x \to 1} \frac{x + x^2 + \dots + x^n - n}{x - 1} = \frac{n(n + 1)}{2}$$ for a given $n$ using l'Hôspital's (Bernoulli's) rule.  The problem is I don't quite like the solution, as it depends on such a heavy weaponry. A limit this simple, should easily be evaluable using some clever idea. Here is a list of what I tried:   Substitute $y = x - 1$. This leads nowhere, I think. Find the Taylor polynomial. Makes no sense, it is a polynomial. Divide by major term. Dividing by $x$ got me nowhere. Find the value $f(x)$ at $x = 1$ directly. I cannot as the function is not defined at $x = 1$. Simplify the expression. I do not see how I could. Using l'Hôspital's (Bernoulli's) rule. Works, but I do not quite like it.   If somebody sees a simple way, please do let me know.    Added later: The approach proposed by Sami Ben Romdhane is universal as asmeurer pointed out. Examples of another limits that can be easily solved this way:   $\lim_{x \to 0} \frac{\sqrt[m]{1 + ax} - \sqrt[n]{1 + bx}}{x}$ where $m, n \in \mathbb{N}$ and $a, b \in \mathbb{R}$ are given, or $\lim_{x \to 0} \frac{\arctan(1 + x) - \arctan(1 - x)}{x}$.   It sems that all limits in the form $\lim_{x \to a} \frac{f(x)}{x - a}$ where $a \in \mathbb{R}$, $f(a) = 0$ and for which $\exists f'(a)$, can be evaluated this way, which is as fast as finding $f'$ and calculating $f'(a)$.  This adds a very useful tool into my calculus toolbox: Some limits can be evaluated easily using derivatives if one looks for $f(a) = 0$, without the l'Hôspital's rule. I have not seen this in widespread use; I propose we call this Sami's rule :).
23626
Evaluate $\displaystyle \lim_{x \to 0} \left(\frac{1}{\ln(1+x)} +  \frac{1}{\ln(1-x)}\right)$   I am having trouble starting this one. I couldn't see any log laws that I'm familiar with to rearrange the formula. I also tried combining the fractions and using L'Hospital's, but it only seemed to make things worse.  What direction should I take with this?
91429
I have a midterm coming up and on the past exams the hard question(s) usually involve some form of $\lim_{x\to0^{+}} x \ln x$. However, we're not allowed to use l'Hopital's rule, on this year's exam anyways.  So how can I evaluate said limit without l'Hopital's rule? I got somewhere with another approach, don't know if it's useful:   $\lim_{x\to0^{+}} x \ln x = \lim_{x\to0^{+}} x^2 \ln (x^2) = L$ $= (\lim_{x\to0^{+}} 2x)(\lim_{x\to0^{+}} x \ln x)$ $= 0 * L$   Then I just need to prove that L is finite/exists (which means it must be 0)
34275
Find the limit $$ \lim_{x \to \infty} \frac{e^{x}}{(1+\frac{1}{x})^{x^{2}}}. $$  I know that the limit is of indeterminate form type $\frac{\infty}{\infty}$, but it seems using L'Hopital's rule directly does not help here.  Do I somehow use the fact that $\lim_{x \to \infty} (1+\frac{1}{x})^{x}=e$?
66681
The limit is $$\lim_{x\to0}\left(\frac{1}{\sin^2x}-\frac{1}{x^2}\right)$$ which I'm aware can be rearranged to obtain the indeterminate $\dfrac{0}{0}$, but in an attempt to avoid L'Hopital's rule (just for fun) I tried using the fact that $\sin x\approx x$ near $x=0$. However, the actual limit is $\dfrac{1}{3}$, not $0$.  In this , the approximation reasoning works out.