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Topics
Pre-Algebra
Mean
Mode
Greatest Common Factor
Least Common Multiple
Order of Operations
Fractions
Mixed Fractions
Prime Factorization
Exponents
Radicals
Algebra
Combine Like Terms
Solve for a Variable
Factor
Expand
Evaluate Fractions
Linear Equations
Quadratic Equations
Inequalities
Systems of Equations
Matrices
Trigonometry
Simplify
Evaluate
Graphs
Solve Equations
Calculus
Derivatives
Integrals
Limits
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\text{Divergent}
$Divergent$
Quiz
Limits
5 problems similar to:
\lim_{ x \rightarrow 0 } \frac{2}{x}
$x→0lim x2 $
Similar Problems from Web Search
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.
Show that Let
$f:R∖{0}→R$
be defined by
$f(x)=x1 $
. Show
$lim_{x→0}x1 $
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", ...
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]}
Find
$lim_{x→0}[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 ...
For
$x→0$
the expression
$[x]x $
is not well defined since for
$0<x<1$
it corresponds to
$0x $
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
Disprove the limit
$lim_{x→0}x1 =5$
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 ...
Given
$ϵ>0$
, we want to find
$δ>0$
such that if
$∣x−0∣=∣x∣<∣δ∣$
then
$∣x1 −5∣<ϵ$
. Of course,
$∣x1 −5∣=∣x1−5x ∣$
so, if x is positive,
$∣x1 −5∣<ϵ$
...
Is this a valid use of l'Hospital's Rule? Can it be used recursively?
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} ...
L'Hôpital's Rule Assuming that the following conditions are true:
$f(x)$
and
$g(x)$
must be differentiable
$dxd g(x)=0$
...
How to explain that division by 0 yields infinity to a 2nd grader
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 ...
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)
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 ...
Some items have been dealt with in comments, so we look only at c). We want to show that for any
$ϵ>0$
, there is a
$B$
such that if
$x>B$
then
$∣sin(1/x)−0∣<ϵ.$
Let
$ϵ>0$
...
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Similar Problems
\lim_{ x \rightarrow 0 } 5
$x→0lim 5$
\lim_{ x \rightarrow 0 } 5x
$x→0lim 5x$
\lim_{ x \rightarrow 0 } \frac{2}{x}
$x→0lim x2 $
\lim_{ x \rightarrow 0 } \frac{1}{x^2}
$x→0lim x_{2}1 $
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