Eduki nagusira salto egin
Microsoft
|
Math Solver
Ebatzi
Praktikatu
Jolastu
Gaiak
Aljebra aurre-álgebra
Esan nahi du
Modua
Zatitzaile komunik handiena
Gutxieneko multiplo komuna
Eragiketen ordena
Zatikiak
Zatiki mistoak
Lehen faktoretza
Adierazleak
Erradikalak
Aljebra
Antzeko terminoak konbinatu
Aldagai bat ebaztea
Faktorea
Zabaldu
Zatikiak ebaluatu
Ekuazio linealak
Ekuazio koadratikoak
Desberdintasunak
Ekuazio-sistemak
Matrizeak
Trigonometria
Sinplifikatu
Ebaluatu
Grafikoak
Ekuazioak ebaztea
Kalkulua
Deribatuak
Integralak
Mugak
Aljebra-sarrerak
Trigonometriako sarrerak
Kalkulu-sarrerak
Matrikula-sarrerak
Ebatzi
Praktikatu
Jolastu
Gaiak
Aljebra aurre-álgebra
Esan nahi du
Modua
Zatitzaile komunik handiena
Gutxieneko multiplo komuna
Eragiketen ordena
Zatikiak
Zatiki mistoak
Lehen faktoretza
Adierazleak
Erradikalak
Aljebra
Antzeko terminoak konbinatu
Aldagai bat ebaztea
Faktorea
Zabaldu
Zatikiak ebaluatu
Ekuazio linealak
Ekuazio koadratikoak
Desberdintasunak
Ekuazio-sistemak
Matrizeak
Trigonometria
Sinplifikatu
Ebaluatu
Grafikoak
Ekuazioak ebaztea
Kalkulua
Deribatuak
Integralak
Mugak
Aljebra-sarrerak
Trigonometriako sarrerak
Kalkulu-sarrerak
Matrikula-sarrerak
Oinarrizko
aljebra
trigonometria
kalkulua
estatistika
Matrizeak
Karaktereak
Ebaluatu
0
Azterketa
Limits
antzeko 5 arazoen antzekoak:
\lim_{ x \rightarrow 0 } 5x
Bilaketaren antzeko arazoak webgunean
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 ...
Gehiago Artikuluak
Partekatu
Kopiatu
Kopiatu portapapeletan
Antzeko arazoak
\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}
Hasierara itzuli
