Ves al contingut principal
Microsoft
|
Math Solver
Resoldre
Pràctica
Jugar
Temes
Preàlgebra
Significar
Moda
Factor comú més gran
Múltiple menys comú
Ordre d'operacions
Fraccions
Fraccions mixtes
Factorització primera:
Exponents
Radicals lliures
Àlgebra
Combina termes semblants
Resoldre per a una variable
Factor
Expandir
Avaluar fraccions
Equacions lineals
Equacions quadràtiques
Desigualtats
Sistemes d'equacions
Matrius
Trigonometria
Simplificar
Avaluar
Gràfics
Resoldre equacions
Càlcul
Derivats
Integrals
Límits
Entrades d'àlgebra
Entrades de trigonometria
Entrades de càlcul
Entrades matricials
Resoldre
Pràctica
Jugar
Temes
Preàlgebra
Significar
Moda
Factor comú més gran
Múltiple menys comú
Ordre d'operacions
Fraccions
Fraccions mixtes
Factorització primera:
Exponents
Radicals lliures
Àlgebra
Combina termes semblants
Resoldre per a una variable
Factor
Expandir
Avaluar fraccions
Equacions lineals
Equacions quadràtiques
Desigualtats
Sistemes d'equacions
Matrius
Trigonometria
Simplificar
Avaluar
Gràfics
Resoldre equacions
Càlcul
Derivats
Integrals
Límits
Entrades d'àlgebra
Entrades de trigonometria
Entrades de càlcul
Entrades matricials
Bàsic
àlgebra
trigonometria
càlcul
estadística
Matrius
Caràcters
Calcula
5
Prova
Limits
5 problemes similars a:
\lim_{ x \rightarrow 0 } 5
Problemes similars de la cerca web
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) ...
Més Elements
Compartir
Copiar
Copiat al porta-retalls
Problemes similars
\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}
Tornar a l'inici
