Langkau ke kandungan utama
Microsoft
|
Math Solver
Selesaikan
Latihan
Bermain
Topik
Pra-algebra
Min
Mod
Faktor Sepunya Terbesar
Gandaan Sepunya Terkecil
Aturan Operasi
Pecahan
Pecahan Campuran
Pemfaktoran Perdana
Eksponen
Radikal
Algebra
Gabungkan Istilah Serupa
Selesaikan untuk mendapat pemboleh ubah
Faktor
Kembangkan
Menilai Pecahan
Persamaan Linear
Persamaan Kuadratik
Ketaksamaan
Sistem Persamaan
Matriks
Trigonometri
Permudahkan
Menilai
Graf
Selesaikan Persamaan
Kalkulus
Derivatif
Kamiran
Had
Input Algebra
Input Trigonometri
Input Kalkulus
Input Matriks
Selesaikan
Latihan
Bermain
Topik
Pra-algebra
Min
Mod
Faktor Sepunya Terbesar
Gandaan Sepunya Terkecil
Aturan Operasi
Pecahan
Pecahan Campuran
Pemfaktoran Perdana
Eksponen
Radikal
Algebra
Gabungkan Istilah Serupa
Selesaikan untuk mendapat pemboleh ubah
Faktor
Kembangkan
Menilai Pecahan
Persamaan Linear
Persamaan Kuadratik
Ketaksamaan
Sistem Persamaan
Matriks
Trigonometri
Permudahkan
Menilai
Graf
Selesaikan Persamaan
Kalkulus
Derivatif
Kamiran
Had
Input Algebra
Input Trigonometri
Input Kalkulus
Input Matriks
Asas
algebra
trigonometri
kalkulus
statistik
matriks
Aksara
Nilaikan
5
Kuiz
Limits
5 masalah yang serupa dengan:
\lim_{ x \rightarrow 0 } 5
Masalah Sama dari Carian Web
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) ...
Lagi Item
Kongsi
Salin
Disalin ke papan klip
Masalah yang serupa
\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}
Kembali ke atas
