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
5
Quiz
Limits
5 ບັນຫາທີ່ຄ້າຍຄືກັນກັບ:
\lim_{ x \rightarrow 0 } 5
ບັນຫາທີ່ຄ້າຍຄືກັນຈາກWeb Search
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) ...
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