Lewati ke konten utama
Microsoft
|
Math Solver
Selesaikan
Berlatih
Bermain
Topik
Pra-Aljabar
Mean
Mode
Faktor Persekutuan Terbesar
Kelipatan Persekutuan Terkecil
Urutan Operasi
Pecahan
Pecahan Campuran
Faktorisasi Prima
Eksponen
Akar
Aljabar
Gabungkan Istilah-Istilah Serupa
Penyelesaian Satu Variabel
Faktor
Ekspansi
Menyelesaikan Pecahan
Persamaan Linear
Persamaan Kuadrat
Ketidaksetaraan
Sistem Persamaan
Matriks
Trigonometri
Menyederhanakan
Menyelesaikan
Grafik
Menyelesaikan Persamaan
Kalkulus
Turunan
Integral
Limit
Input Aljabar
Input Trigonometri
Input Kalkulus
Input Matriks
Selesaikan
Berlatih
Bermain
Topik
Pra-Aljabar
Mean
Mode
Faktor Persekutuan Terbesar
Kelipatan Persekutuan Terkecil
Urutan Operasi
Pecahan
Pecahan Campuran
Faktorisasi Prima
Eksponen
Akar
Aljabar
Gabungkan Istilah-Istilah Serupa
Penyelesaian Satu Variabel
Faktor
Ekspansi
Menyelesaikan Pecahan
Persamaan Linear
Persamaan Kuadrat
Ketidaksetaraan
Sistem Persamaan
Matriks
Trigonometri
Menyederhanakan
Menyelesaikan
Grafik
Menyelesaikan Persamaan
Kalkulus
Turunan
Integral
Limit
Input Aljabar
Input Trigonometri
Input Kalkulus
Input Matriks
Dasar
Aljabar
trigonometri
Kalkulus
statistik
Matriks
Karakter
Evaluasi
0
Kuis
Limits
5 soal serupa dengan:
\lim_{ x \rightarrow 0 } 5x
Soal yang Mirip dari Pencarian Web
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 ...
Lebih banyak Item
Bagikan
Salin
Disalin ke clipboard
Masalah 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
