Skip ໄປຫາເນື້ອຫາຫຼັກ
Microsoft
|
Math Solver
ແກ້
ການປະຕິບັດ
ຫຼິ້ນ
ຫົວຂໍ້
Pre-Algebra
Mean
Mode
ປັດໄຈທົ່ວໄປທີ່ຍິ່ງໃຫຍ່ທີ່ສຸດ
ຫນ້ອຍທີ່ສຸດທົ່ວໄປຫຼາຍ
ລະບຽບການດໍາເນີນງານ
ສ່ວນຍ່ອຍ
ສ່ວນປະກອບປະສົມ
ປັດຈຸບັນ
Exponents
Radicals
Algebra
Combine Like Terms
ແກ້ໄຂສໍາລັບVariable
ປັດໃຈ
ຂະຫຍາຍ
ປະເມີນຜົນສ່ວນປະກອບ
ສະສົມLinear Equations
ສະສົມQuadratic
ຄວາມບໍ່ສະເຫມີພາບ
ລະບົບຂອງEquations
ມັດທຣີສ
Trigonometry
ລຽບງ່າຍ
ປະເມີນຜົນ
Graphs
ແກ້ໄຂສະສົມ
Calculus
ຜະລິດຕະພັນ
Integrals
ຂີດຈໍາກັດ
Algebra Inputs
Trigonometry Inputs
Calculus Inputs
Matrix Inputs
ແກ້
ການປະຕິບັດ
ຫຼິ້ນ
ຫົວຂໍ້
Pre-Algebra
Mean
Mode
ປັດໄຈທົ່ວໄປທີ່ຍິ່ງໃຫຍ່ທີ່ສຸດ
ຫນ້ອຍທີ່ສຸດທົ່ວໄປຫຼາຍ
ລະບຽບການດໍາເນີນງານ
ສ່ວນຍ່ອຍ
ສ່ວນປະກອບປະສົມ
ປັດຈຸບັນ
Exponents
Radicals
Algebra
Combine Like Terms
ແກ້ໄຂສໍາລັບVariable
ປັດໃຈ
ຂະຫຍາຍ
ປະເມີນຜົນສ່ວນປະກອບ
ສະສົມLinear Equations
ສະສົມQuadratic
ຄວາມບໍ່ສະເຫມີພາບ
ລະບົບຂອງEquations
ມັດທຣີສ
Trigonometry
ລຽບງ່າຍ
ປະເມີນຜົນ
Graphs
ແກ້ໄຂສະສົມ
Calculus
ຜະລິດຕະພັນ
Integrals
ຂີດຈໍາກັດ
Algebra Inputs
Trigonometry Inputs
Calculus Inputs
Matrix Inputs
ພື້ນຖານ
algebra
trigonometry
calculus
ສະຖິຕິ
matrices
ຕົວອັກສອນ
ປະເມີນ
0
Quiz
Limits
5 ບັນຫາທີ່ຄ້າຍຄືກັນກັບ:
\lim_{ x \rightarrow 0 } 5x
ບັນຫາທີ່ຄ້າຍຄືກັນຈາກWeb Search
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 ...
ລາຍການ
ແບ່ງປັນ
ສໍາເນົາ
ສໍາເນົາຄລິບ
ບັນຫາທີ່ຄ້າຍຄືກັນ
\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}
ກັບຄືນສູ່ຍອດ
