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
\text{Divergent}
Kuis
Limits
5 soal serupa dengan:
\lim_{ x \rightarrow 0 } \frac{2}{x}
Soal yang Mirip dari Pencarian Web
Show that Let f : \mathbb{R} \setminus \{0\} \to \mathbb{R} be defined by f(x) = \frac{1}{x}. Show \lim_{x \to 0}\frac{1}{x} doesn't exist.
https://math.stackexchange.com/q/2826102
Suppose that f: U → R is an application defined on a subset U of the set R of reals. If p is a real, not necessarily belonging to U but such that f is "defined in the neighborhood of p", ...
Find \lim_{x\rightarrow0}\frac{x}{[x]}
https://math.stackexchange.com/q/2835948
For x\to 0 the expression \frac{x}{[x]} is not well defined since for 0<x<1 it corresponds to \frac x 0 and thus we can't calculate the limit for that expression. As you noticed, we can only ...
Disprove the limit \lim_{x\to 0}\frac{1}{x}=5 with epsilon-delta
https://math.stackexchange.com/q/1527181
Given \epsilon> 0, we want to find \delta> 0 such that if |x- 0|= |x|< |\delta| then |\frac{1}{x}- 5|< \epsilon. Of course, |\frac{1}{x}- 5|= |\frac{1- 5x}{x}| so, if x is positive, |\frac{1}{x}- 5|<\epsilon ...
Is this a valid use of l'Hospital's Rule? Can it be used recursively?
https://math.stackexchange.com/questions/946785/is-this-a-valid-use-of-lhospitals-rule-can-it-be-used-recursively
L'Hôpital's Rule Assuming that the following conditions are true: f(x) and g(x) must be differentiable \frac{d}{dx}g(x)\neq 0 \lim\limits_{x\to c} \frac{f(x)}{g(x)}= \frac{0}{0}\mbox{ or }\lim\limits_{x\to c} \frac{f(x)}{g(x)}= \frac{\pm\infty}{\pm\infty} ...
How to explain that division by 0 yields infinity to a 2nd grader
https://math.stackexchange.com/questions/242258/how-to-explain-that-division-by-0-yields-infinity-to-a-2nd-grader
The first thing to point out is that division by zero is not defined! You cannot divide by zero. Consider the number 1/x where x is a negative number. You will find that 1/x is negative for all ...
precise definition of a limit at infinity, application for limit at sin(x)
https://math.stackexchange.com/questions/1776133/precise-definition-of-a-limit-at-infinity-application-for-limit-at-sinx
Some items have been dealt with in comments, so we look only at c). We want to show that for any \epsilon\gt 0, there is a B such that if x\gt B then |\sin(1/x)-0|\lt \epsilon. Let \epsilon\gt 0 ...
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