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
0
Prova
Limits
5 problemes similars a:
\lim_{ x \rightarrow 0 } 5x
Problemes similars de la cerca web
Prove that for any c \neq 0 \lim_{x \rightarrow c}{h(x)} does not exist and that \lim_{x \rightarrow 0}{h(x)} does exist.
https://math.stackexchange.com/questions/334631/prove-that-for-any-c-neq-0-lim-x-rightarrow-chx-does-not-exist-and
Hint: take one sequence that contains only rationals and another one that contains only irrationals (both tending to c\ne 0). For the case of c=0, you can use e.g. that h is continuous at 0 ...
Proofs regarding Continuous functions 1
https://math.stackexchange.com/questions/526691/proofs-regarding-continuous-functions-1
The proof of part a) needs to be modified a bit. You have used the logic that if N \leq f(x) \leq M then xN \leq xf(x) \leq xM. This holds only when x \geq 0. It is better to change the argument ...
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) ...
Calculate: \lim_{x \to 0 } = x \cdot \sin(\frac{1}{x})
https://math.stackexchange.com/questions/1066434/calculate-lim-x-to-0-x-cdot-sin-frac1x
Your proof is incorrect, cause you used incorrect transform, but it has already been stated. I'll describe way to solve it. \lim_{x \to 0}\frac{\sin(\frac{1}{x})}{\frac{1}{x}} \neq 1 Hint : ...
Prove that f(x) is bounded. Please check my proof.
https://math.stackexchange.com/q/1052420
Here is another approach: Let L_0 = \lim_{x \downarrow 0} f(x), L_\infty = \lim_{x \to \infty} f(x). By definition of the limit we have some \delta>0 and N>0 such that if x \in (0, \delta), ...
Complex Function limit by investigating sequences
https://math.stackexchange.com/questions/1915934/complex-function-limit-by-investigating-sequences
If a limit as z \to 0 exists, one should be able to plug in any sequence \{ z_n \} going to zero and get the same limit. Limits of sequences are generally easier to work with. So in this case if ...
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