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.

2796 questions
522
votes
27 answers

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
  • 8,649
  • 5
  • 27
  • 47
115
votes
3 answers

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
  • 268,487
  • 17
  • 196
  • 314
51
votes
7 answers

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
  • 511
  • 5
  • 3
28
votes
9 answers

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
  • 423
  • 2
  • 5
  • 15
26
votes
5 answers

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
  • 269
  • 3
  • 3
23
votes
11 answers

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
  • 1,913
  • 9
  • 22
20
votes
5 answers

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
  • 15,266
  • 2
  • 45
  • 104
19
votes
5 answers

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
  • 5,279
  • 2
  • 22
  • 64
19
votes
8 answers

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
  • 323
  • 2
  • 6
18
votes
8 answers

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
  • 1,453
  • 11
  • 20
18
votes
1 answer

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
  • 4,992
  • 2
  • 18
  • 46
17
votes
7 answers

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
  • 841
  • 1
  • 4
  • 18
16
votes
7 answers

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
  • 161
  • 3
16
votes
5 answers

Evaluating $\lim_{x\to 0}\left(\frac{1}{\sin x} - \frac{1}{\tan x}\right)$

How to solve this limit $$ \lim_{x\to 0}\left(\frac{1}{\sin x} - \frac{1}{\tan x}\right) $$ without using L'Hospital's rule?
whyguy
  • 323
  • 1
  • 9
16
votes
2 answers

How do I calculate this limit: $\lim\limits_{n\to\infty}1+\sqrt[2]{2+\sqrt[3]{3+\dotsb+\sqrt[n]n}}$?

I have seen this question on the internet and was interested to know the answer. Here it is : Calculate $\lim\limits_{n\to\infty}(1+\sqrt[2]{2+\sqrt[3]{3+\dotsb+\sqrt[n]n}})$? Edit : I really tried doing it but wasn't able to get somewhere. I know…
Sudhanshu
  • 1,497
  • 12
  • 21
1
2 3
…
99 100