Hopp til hovedinnhold
Microsoft
|
Math Solver
Løse
Praksis
Skuespill
Emner
Pre-Algebra
Bety
Modus
Største felles faktor
Minst vanlige multiplum
Rekkefølge av operasjoner
Fraksjoner
Blandede brøker
Førsteklasses faktorisering
Eksponenter
Radikaler
Algebra
Kombiner som termer
Løse for en variabel
Faktor
Utvide
Vurdere brøker
Lineære formler
Kvadratiske ligninger
Ulikheter
Ligningssystemer
Matriser
Trigonometri
Forenkle
Vurdere
Grafer
Løs formler
Beregning
Derivater
Integraler
Grenser
Algebra innganger
Trigonometri-innganger
Kalkulus innganger
Matrise innganger
Løse
Praksis
Skuespill
Emner
Pre-Algebra
Bety
Modus
Største felles faktor
Minst vanlige multiplum
Rekkefølge av operasjoner
Fraksjoner
Blandede brøker
Førsteklasses faktorisering
Eksponenter
Radikaler
Algebra
Kombiner som termer
Løse for en variabel
Faktor
Utvide
Vurdere brøker
Lineære formler
Kvadratiske ligninger
Ulikheter
Ligningssystemer
Matriser
Trigonometri
Forenkle
Vurdere
Grafer
Løs formler
Beregning
Derivater
Integraler
Grenser
Algebra innganger
Trigonometri-innganger
Kalkulus innganger
Matrise innganger
Grunnleggende
algebra
trigonometri
beregning
statistikk
Matriser
Tegn
Evaluer
5
Spørrelek
Limits
5 problemer som ligner på:
\lim_{ x \rightarrow 0 } 5
Lignende problemer fra nettsøk
Is \lim_{x\to 0} (x) different from dx
https://math.stackexchange.com/questions/1157952/is-lim-x-to-0-x-different-from-dx
It is confusing because the way derivatives are taught today are different from how it was done back in the 1600s. Back then a derivative was dy/dx, where dy and dx were infinitesimal ...
Calculating the limit: \lim \limits_{x \to 0} \frac{\ln(\frac{\sin x}{x})}{x^2}.
https://math.stackexchange.com/q/1147074
We want L = \lim_{x\to 0} \frac{\ln(\frac{\sin x}{x})}{x^2} Since the top approaches \ln(1) = 0 and the bottom also approaches 0, we may use L'Hopital: L = \lim_{x\to 0}{\frac{(\frac{x}{\sin x})(\frac{x \cos x - \sin x}{x^2})}{2x}} = \lim_{x\to 0}\frac{x \cos x - \sin x}{2x^2\sin x} ...
Left/right-hand limits and the l'Hôpital's rule
https://math.stackexchange.com/q/346759
In this very case it is even simpler: the limit (not one sided!) exists, so you don't even need to split the calculation in two steps! And yes: apply l'Hospital directly to the limit .
Arrow in limit operator
https://math.stackexchange.com/questions/36333/arrow-in-limit-operator
Yes, it means that considers decreasing sequences that converge to 0. I've only once worked with someone who preferred to use the \searrow and \nearrow notation, but it's a good notation in the ...
Prob. 15, Sec. 5.1, in Bartle & Sherbert's INTRO TO REAL ANALYSIS: A bounded function on (0, 1) having no limit as x \to 0
https://math.stackexchange.com/q/2879789
What you did is correct. In order to show that \alpha\neq\beta, suppose otherwise. That is, suppose that \alpha=\beta. I will prove that \lim_{x\to0}f(x)=\alpha(=\beta), thereby reaching a ...
Use L'Hopital's with this problem?
https://math.stackexchange.com/questions/1419122/use-lhopitals-with-this-problem
Let \displaystyle y=\lim_{x\rightarrow 0^{+}}\left(\frac{1}{x}\right)^{\sin x}\;, Now Let x=0+h\;, Then \displaystyle y=\lim_{h\rightarrow 0}\left(\frac{1}{h}\right)^{\sin h} So \displaystyle \ln(y) = \lim_{h\rightarrow 0}\sin (h)\cdot \ln\left(\frac{1}{h}\right) = -\lim_{h\rightarrow 0}\sin h\cdot \ln(h) = -\lim_{h\rightarrow 0}\frac{\ln(h)}{\csc (h)}\left(\frac{\infty}{\infty}\right) ...
Flere Elementer
Aksje
Kopi
Kopiert til utklippstavle
Lignende problemer
\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}
Tilbake til toppen