Tīpoka ki ngā ihirangi matua
Microsoft
|
Math Solver
Whakatau
Whakaharatau
Tākaro
Ngā Kaupapa
Pre-Algebra
Mean
Aratau:
Āhuatanga Noa Nui Rawa
He maha rawa ngā mea noa iho
Raupapa Mahi
Ngā Hautanga
Ngā Hautanga Whāranu
Āhuatanga Matua
Ngā Exponents
Ngā Radicals
Algebra
Paheko pēnei i ngā Ture
Whakaoti mō tētahi Tāupe
Āhuatanga
Whakaroha
Evaluate Fractions
Whārite Paerangi
Ngā Whārite Tapawhā
Ōritetanga
Ngā Pūnaha Whārite
Matrices
Āhuahanga
Whakangāwari
Evaluate
Ngā Graphs
Whakatau Whārite
Tātaitai
Ngā Āhuatanga
Integrals
Ngā Tepe
Ngā Tāuru Algebra
Ngā Tāuru Āhuahanga
Ngā Tāuru Tātai
Ngā Tāuru Poukapa
Whakatau
Whakaharatau
Tākaro
Ngā Kaupapa
Pre-Algebra
Mean
Aratau:
Āhuatanga Noa Nui Rawa
He maha rawa ngā mea noa iho
Raupapa Mahi
Ngā Hautanga
Ngā Hautanga Whāranu
Āhuatanga Matua
Ngā Exponents
Ngā Radicals
Algebra
Paheko pēnei i ngā Ture
Whakaoti mō tētahi Tāupe
Āhuatanga
Whakaroha
Evaluate Fractions
Whārite Paerangi
Ngā Whārite Tapawhā
Ōritetanga
Ngā Pūnaha Whārite
Matrices
Āhuahanga
Whakangāwari
Evaluate
Ngā Graphs
Whakatau Whārite
Tātaitai
Ngā Āhuatanga
Integrals
Ngā Tepe
Ngā Tāuru Algebra
Ngā Tāuru Āhuahanga
Ngā Tāuru Tātai
Ngā Tāuru Poukapa
Taketake
papara
ahuatoru
tatau
Ngā tatauranga
matrices
Ngā Pūāhua
Aromātai
\text{Divergent}
Pātaitai
Limits
5 raruraru e ōrite ana ki:
\lim_{ x \rightarrow 0 } \frac{2}{x}
Ngā Raru Ōrite mai i te Rapu Tukutuku
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 ...
Ētahi atu Ngā tūemi
Tohaina
Tārua
Kua tāruatia ki te papatopenga
Ngā Raru Ōrite
\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}
Hoki ki runga
