If the expression obtained after any substitution during limit analysis does not give enough information to determine the original limit, it is known as an indeterminate form.

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computing an indeterminate form of a limit

I need a hint on computing this one limit: $$\lim_{n \to\infty}\,\frac{2\cdot4^n+3\cdot n^4}{4\cdot n^6-3\cdot 3^n +3n}$$ Thank you
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Limit Problem : Infinity Limit Problem

I can't find the way to solve this question and i always get 0/0. The question is: $\lim_{x\rightarrow \infty } x\left [ 2^{\frac{1}{x}}-1 \right ]$ From Mathematica, i get -infinity. But how can it ...
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39 views

What is the value of 0/0? [duplicate]

I have heard that anything divided by zero is infinity, so i was wondering what would be 0/0? I am a high school student and I haven't studied calculus yet.
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Why Is $1^{\infty}$ an indeteminate form? [duplicate]

For me, multiplication is a binary operation so it can be applied only on a finite sequence of numbers. but $1^{\infty}$ requires that we apply multiplication infinitly which is not defined as ...
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why is $\displaystyle \frac{\log(\sin x)}{\log(x)}$ $\quad\frac{\infty}{\infty}$ form as $x\to 0$?

In this question $\displaystyle\frac{\log(\sin x)}{\log x}$ is taken as $\displaystyle\frac{\infty}{\infty}$ indeterminate form. But $\log(0)$ is not defined so how can L'Hospital's rule can be ...
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Why is this rational expressions indeterminate when evaluated?

I have this rational expression to evaluate, $$ {{3a-3}\over {4a(a-1)}} \text { if } a=1. $$ I understand that if you substitute 1, both the numerator and denominator would turn out 0, thus making ...
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248 views

computing a limit of a function that is positive defined

given a function $f(x)$ positive and continuous at $x=a$, with $f(a)\ne0$, compute: $$\lim_{n\to+\infty}\left[\frac{f\left(a+\frac{1}{n}\right)}{f(a)}\right]^n$$ a friend asked me that, though I ...
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3answers
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indeterminate limit where applying L'Hopitals Rules directly doesn't help and using ln gives wrong answer

I am trying to determine the limit $\displaystyle{ \lim_{x \to 0^-}{\frac{-e^{1/x}}{x}}}$. Plugging in $x$ directly, yields $0/0$ which is indeterminate. Applying L'Hopitals rule does not simplify ...
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I got the answer for $\lim \limits_{x \to \infty} {\left({3x-2 \over3x+4}\right)}^{3x+1}$, but only by a mistake - how do I solve correctly?

This is what I did for: $$\lim \limits_{x \to \infty} {\left({3x-2 \over3x+4}\right)}^{3x+1}$$ Check form: $\left({\infty \over \infty}\right)^{\infty}$. Apply L'Hospital's Rule to just $\lim ...
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2answers
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Easier way to solve $\lim \limits_{x \to 0} \space \cot{x}-{1 \over x}$ using L'Hospital's Rule?

This is what I did for: $$\lim \limits_{x \to 0} \space \cot{x}-{1 \over x}$$ Check form: $\infty - \infty$. Rearrange it to be a quotient: $$\begin{align} \\ & =\lim \limits_{x \to 0} \space ...
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Is $(-1)^{\infty}$ an indeterminate form?

We know that $\lim_{n\to\infty}(-1)^n=(-1)^{\infty}$ doesn't exist. Now take $\lim_{n\to\infty}(-1)^{2n}=(-1)^{\infty}$. This limit exists, because ...
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1answer
91 views

Determination of $1^\infty$ indeterminate forms

Recently I have been learning some of the basic concepts of limits and in my academics. There I have been taught some methods to evaluate indeterminant forms like $1^\infty$, $0^0$ and $\infty^0$. ...
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2answers
71 views

How to explain indeterminations, and some aprpoaches to $+\infty$ or $-\infty$, for middle school students?

Question: how to explain the undefinitions $0^0$ and $\frac{0}{0}$ for Middle school students?? I am a math teacher and I don't know how to answer properly when studens ask me why some operations ...
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3answers
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Using L'Hospital's Rule to evaluate limit to infinity

I'm given this problem and I'm not sure how to solve it. I was only ever given one example in class on using L'Hospital's rule like this, but it is very different from this particular problem. Can ...
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1answer
70 views

Help me integrate this function using Simpson's rule

I have a question: compute $$\int_0^1 \frac{\sin(x)}{x}\,dx$$ for $n=10$ divisions. I got the value $0.9127$ but I think its a bit too high.
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Why can $2^3$ be defined but $0^0$ cannot

From what I gather, we can't just define $0^0$ to be $0$ or $1$ or $69$ or whatever, because $\lim\limits_{x\mathop\to0}0^x=0$ and $\lim\limits_{x\mathop\to0}x^0=1$. So $0^0$ is called indeterminate ...
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Limits of Indeterminate Powers in Exponential Form using L'Hopital's Rule

I am trying to find the limit as $x \rightarrow 0$ of $x^x$ using L'Hopital's rule. I have written it in exponential form: $\lim\limits_{x \rightarrow 0} e^{x \ln x}$. I do not know how to put it in ...
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3answers
41 views

Limit of a polynomic-exponential sequence

I have to calculate the following limit: $$L=\lim \limits_{n \to \infty} -(n-n^{n/(1+n)})$$ I get the indeterminate form $\infty - \infty$ and I don't know how to follow. Any idea? Thank you very ...
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Indeterminate limit that is supposed to be solved with De L'Hospital's rule

Last week my Maths teacher gave the class this exercise taken from our text book. We are working on De L'Hospital's rule at the moment and this exercise is from that part of the book so everybody ...
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Is zero a singular point of this function?

$$f(z)=\frac{z^3}{z+z^5}$$I thought that this function has 5 singular points. But my friend is convinced it only has four because if you write is as$$f(z)=\frac{z^2}{1+z^4}$$ then it is defined at ...
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Solving limit without L'Hopital

I'd like some help in solving this limit without using L'Hopital. $$\lim_{x\to -\infty}\frac{\ln(1-2x)}{1-\sqrt{1-x}}$$ I've also solved it changing the variable to $y=\sqrt{1-x}$ but I would like ...
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Is it wrong to tell children that 1/0 = NaN is incorrect, and should be ∞?

I was on the tube and overheard a dad questioning his kids about maths. The children were probably about 11 or 12 years old. After several more mundane questions he asked his daughter what 1/0 ...