Overslaan en naar de inhoud gaan
Microsoft
|
Math Solver
Oplossen
Oefenen
Spelen
Onderwerpen
Pre-Algebra
Gemiddelde
Modus
Grootste Gemene Deler
Kleinste Gemene Veelvoud
Bewerkingsvolgorde
Breuken
Gemengde Breuken
Ontbinding in priemfactoren
Exponenten
Wortels
Algebra
Combineer Soortgelijke Termen
Een variabele oplossen
Factor
Uitbreiden
Breuken evalueren
Lineaire Vergelijkingen
Vierkantsvergelijkingen
Ongelijkheden
Stelsels van vergelijking
Matrices
Trigonometrie
Vereenvoudigen
Evalueren
Grafieken
Vergelijkingen oplossen
Analyse
Afgeleiden
Integralen
Limieten
Algebra-ingangen
Trigonometrische ingangen
Calculus-invoer
Matrix-ingangen
Oplossen
Oefenen
Spelen
Onderwerpen
Pre-Algebra
Gemiddelde
Modus
Grootste Gemene Deler
Kleinste Gemene Veelvoud
Bewerkingsvolgorde
Breuken
Gemengde Breuken
Ontbinding in priemfactoren
Exponenten
Wortels
Algebra
Combineer Soortgelijke Termen
Een variabele oplossen
Factor
Uitbreiden
Breuken evalueren
Lineaire Vergelijkingen
Vierkantsvergelijkingen
Ongelijkheden
Stelsels van vergelijking
Matrices
Trigonometrie
Vereenvoudigen
Evalueren
Grafieken
Vergelijkingen oplossen
Analyse
Afgeleiden
Integralen
Limieten
Algebra-ingangen
Trigonometrische ingangen
Calculus-invoer
Matrix-ingangen
Basic
algebra
Trigonometrie
analyse
statistieken
matrices
Karakters
Evalueren
\text{Divergent}
Quiz
Limits
\lim_{ x \rightarrow 0 } \frac{2}{x}
Vergelijkbare problemen van Web Search
Show that Let f : \mathbb{R} \setminus \{0\} \to \mathbb{R} be defined by f(x) = \frac{1}{x}. Show \lim_{x \to 0}\frac{1}{x} doesn't exist.
https://math.stackexchange.com/q/2826102
Suppose that f: U → R is an application defined on a subset U of the set R of reals. If p is a real, not necessarily belonging to U but such that f is "defined in the neighborhood of p", ...
Find \lim_{x\rightarrow0}\frac{x}{[x]}
https://math.stackexchange.com/q/2835948
For x\to 0 the expression \frac{x}{[x]} is not well defined since for 0<x<1 it corresponds to \frac x 0 and thus we can't calculate the limit for that expression. As you noticed, we can only ...
Disprove the limit \lim_{x\to 0}\frac{1}{x}=5 with epsilon-delta
https://math.stackexchange.com/q/1527181
Given \epsilon> 0, we want to find \delta> 0 such that if |x- 0|= |x|< |\delta| then |\frac{1}{x}- 5|< \epsilon. Of course, |\frac{1}{x}- 5|= |\frac{1- 5x}{x}| so, if x is positive, |\frac{1}{x}- 5|<\epsilon ...
Is this a valid use of l'Hospital's Rule? Can it be used recursively?
https://math.stackexchange.com/questions/946785/is-this-a-valid-use-of-lhospitals-rule-can-it-be-used-recursively
L'Hôpital's Rule Assuming that the following conditions are true: f(x) and g(x) must be differentiable \frac{d}{dx}g(x)\neq 0 \lim\limits_{x\to c} \frac{f(x)}{g(x)}= \frac{0}{0}\mbox{ or }\lim\limits_{x\to c} \frac{f(x)}{g(x)}= \frac{\pm\infty}{\pm\infty} ...
How to explain that division by 0 yields infinity to a 2nd grader
https://math.stackexchange.com/questions/242258/how-to-explain-that-division-by-0-yields-infinity-to-a-2nd-grader
The first thing to point out is that division by zero is not defined! You cannot divide by zero. Consider the number 1/x where x is a negative number. You will find that 1/x is negative for all ...
precise definition of a limit at infinity, application for limit at sin(x)
https://math.stackexchange.com/questions/1776133/precise-definition-of-a-limit-at-infinity-application-for-limit-at-sinx
Some items have been dealt with in comments, so we look only at c). We want to show that for any \epsilon\gt 0, there is a B such that if x\gt B then |\sin(1/x)-0|\lt \epsilon. Let \epsilon\gt 0 ...
Meer Items
Delen
Kopiëren
Gekopieerd naar klembord
Soortgelijke problemen
\lim_{ x \rightarrow 0 } 5
\lim_{ x \rightarrow 0 } 5x
\lim_{ x \rightarrow 0 } \frac{2}{x}
\lim_{ x \rightarrow 0 } \frac{1}{x^2}
Terug naar boven