Tīpoka ki ngā ihirangi matua
Microsoft
|
Math Solver
Whakatau
Whakaharatau
Tākaro
Ngā Kaupapa
Pre-Algebra
Mean
Aratau:
Āhuatanga Noa Nui Rawa
He maha rawa ngā mea noa iho
Raupapa Mahi
Ngā Hautanga
Ngā Hautanga Whāranu
Āhuatanga Matua
Ngā Exponents
Ngā Radicals
Algebra
Paheko pēnei i ngā Ture
Whakaoti mō tētahi Tāupe
Āhuatanga
Whakaroha
Evaluate Fractions
Whārite Paerangi
Ngā Whārite Tapawhā
Ōritetanga
Ngā Pūnaha Whārite
Matrices
Āhuahanga
Whakangāwari
Evaluate
Ngā Graphs
Whakatau Whārite
Tātaitai
Ngā Āhuatanga
Integrals
Ngā Tepe
Ngā Tāuru Algebra
Ngā Tāuru Āhuahanga
Ngā Tāuru Tātai
Ngā Tāuru Poukapa
Whakatau
Whakaharatau
Tākaro
Ngā Kaupapa
Pre-Algebra
Mean
Aratau:
Āhuatanga Noa Nui Rawa
He maha rawa ngā mea noa iho
Raupapa Mahi
Ngā Hautanga
Ngā Hautanga Whāranu
Āhuatanga Matua
Ngā Exponents
Ngā Radicals
Algebra
Paheko pēnei i ngā Ture
Whakaoti mō tētahi Tāupe
Āhuatanga
Whakaroha
Evaluate Fractions
Whārite Paerangi
Ngā Whārite Tapawhā
Ōritetanga
Ngā Pūnaha Whārite
Matrices
Āhuahanga
Whakangāwari
Evaluate
Ngā Graphs
Whakatau Whārite
Tātaitai
Ngā Āhuatanga
Integrals
Ngā Tepe
Ngā Tāuru Algebra
Ngā Tāuru Āhuahanga
Ngā Tāuru Tātai
Ngā Tāuru Poukapa
Taketake
papara
ahuatoru
tatau
Ngā tatauranga
matrices
Ngā Pūāhua
Aromātai
0
Pātaitai
Limits
5 raruraru e ōrite ana ki:
\lim_{ x \rightarrow 0 } 5x
Ngā Raru Ōrite mai i te Rapu Tukutuku
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 ...
Ētahi atu Ngā tūemi
Tohaina
Tārua
Kua tāruatia ki te papatopenga
Ngā Raru Ōrite
\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}
Hoki ki runga