Mathematics Stack Exchange
Mathematics Stack Exchange

Questions tagged [limits-without-lhopital]

The evaluation of limits without the usage of L'Hôpital's rule.

The idea here is to evaluate the limit using standard limit theorems (algebra of limits, Sandwich/Squeeze Theorem, essentially without using any differentiation) and some standard limit formulas related to algebraic, trigonometric, exponential and logarithmic functions. Very often, Taylor series techniques prove fruitful in such problems as they allow for easy cancellation of powers and most terms evaluate to zero, leaving a simple expression for the limit.

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How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?

How can one prove the statement $$\lim_{x\to 0}\frac{\sin x}x=1$$ without using the Taylor series of $\sin$, $\cos$ and $\tan$? Best would be a geometrical solution. This is homework. In my math class, we are about to prove that $\sin$ is…
FUZxxl
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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…
lab bhattacharjee
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What's wrong with l'Hopital's rule?

Upon looking at yet another question on this site on evaluating a limit explicitly without l'Hopital's rule, I remembered that one of my professors once said something to the effect that in Europe (where he is from) l'Hopital's rule isn't "overused"…
user241336
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Are there any situations in which L'Hopital's Rule WILL NOT work?

Today was my first day learning L'Hopital's Rule, and I was wondering if there are any situations in which you cannot use this rule, with the exception of when a limit is determinable.
ErinAjello
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What is an intuitive approach to solving $\lim_{n\rightarrow\infty}\biggl(\frac{1}{n^2} + \frac{2}{n^2} + \frac{3}{n^2}+\dots+\frac{n}{n^2}\biggr)$?

$$\lim_{n\rightarrow\infty}\biggl(\frac{1}{n^2} + \frac{2}{n^2} + \frac{3}{n^2}+\dots+\frac{n}{n^2}\biggr)$$ I managed to get the answer as $1$ by standard methods of solving, learned from teachers, but my intuition says that the denominator of…
mathlover
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Limit question - L'Hopital's rule doesn't seem to work

I have been recently trying to solve this limit problem. First of all, I used L'Hopital's rule but it doesn't seem to work (because I thought that this limit is of form $\frac{\infty}{\infty}$). Am I doing it correctly? I don't seem to understand…
Akshay Yadav
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Evaluate $\lim_{x\to 0}\frac {(\cos(x))^{\sin(x)} - \sqrt{1 - x^3}}{x^6}$

Evaluate $$ \displaystyle \lim_{x\to 0}\Bigg( \frac {(\cos(x))^{\sin(x)} - \sqrt{1 - x^3}}{x^6}\Bigg).$$ I tried to use L'Hopital's rule but it got very messy. Moreover I also tried to analyze from graphs, but I was getting the limit $= 0$ by…
Henry
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Show $\lim_{h\to 0} \frac{(a^h-1)}{h}$ exists without l'Hôpital or even referencing $e$ or natural log

Taking as our definition of exponentiation repeated multiplication (extended to real exponents by continuity), can we show that the limit $$\lim_{h\to 0}\dfrac{a^h-1}{h}$$ exists, without l'Hôpital, $e$, or even natural logarithm? Sure, l'Hôpital…
ziggurism
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Are there any limit questions which are easier to solve using methods other than l'Hopital's Rule?

Are there any limit questions which are easier to solve using methods other than l'Hopital's Rule? It seems like for every limit that results in an indeterminate form, you may as well use lHopital's Rule rather than any of the methods that are…
user291130
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Why do we need to check for more than $\frac{\infty}{\infty}$ or $\frac{0}{0}$ when applying L'Hospital?

With regard to this comment I wanted to ask (and provide an answer): what else do we need to assume and check for before we can apply L'Hospital's rule? And why do we have to do this e.g. does it really matter if we don't check them?
Hirshy
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With $x_1=1$ and $x_{n+1} = \frac{1}{x_1^2+x_2^2+\dots+x_{n}^2}$show$\lim\limits_{n\to\infty}\frac{x_1+x_2+\dots+x_n}{n^{2/3}}=\frac{\sqrt[3]{9}}{2}$

Context I found an extra-hard (but fake) example of a paper which adheres to the syllabus for the hardest mathematics high school course available in the state of NSW in Australia; Mathematics Extension 2. Question 16 (a) has caught my eye and I'm…
Juan Reight Deag
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Finding the Limit of a Sequence (No L'Hopital's Rule)

Okay, I feel almost silly for asking this, but I've been on it for a good hour and a half. I need to find: $$\lim_{n \to\infty}\left(1-\frac{1}{n^{2}}\right)^{n}$$ But I just can't seem to figure it out. I know its pretty easy using L'Hopital's…
CoffeeCrow
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Understand how to evaluate $\lim _{x\to 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$

We are given this limit to evaluate: $$\lim _{x\to 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$$ In this example, if we try substitution it will lead to an indeterminate form $\frac{0}{0}$. So, in order to evaluate this limit, we can multiply this…
欲しい未来
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Why doesn't using the approximation $\sin x\approx x$ near $0$ work for computing this limit?

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…
user170231
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Finding $\lim_{n \to \infty }\sqrt[n]{b^{2^{-n}}-1}$ without L'hopital

I found the limit $\lim_{n \to \infty }\sqrt[n]{b^{2^{-n}}-1}$ by first defining $f(x)=\sqrt[x]{b^{2^{-x}}-1}$ above $R$ and then finding the limit of $ln(f)$ (to cancel the nth root). This worked (the result is $1/2$), but I ended up having to find…
roy
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