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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
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5
$5$
Quiz
Limits
5 problems similar to:
\lim_{ x \rightarrow 0 } 5
$x→0lim 5$
Similar Problems from Web Search
Is \lim_{x\to 0} (x) different from dx
Is
$lim_{x→0}(x)$
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 ...
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}.
Calculating the limit:
$x→0lim $
$x_{2}ln(xsinx ) .$
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} ...
We want
$L=lim_{x→0}x_{2}ln(xsinx ) $
Since the top approaches
$ln(1)=0$
and the bottom also approaches
$0$
, we may use L'Hopital:
$L=lim_{x→0}2x(sinxx )(xxcosx−sinx ) =lim_{x→0}2x_{2}sinxxcosx−sinx $
...
Left/right-hand limits and the l'Hôpital's rule
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 .
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
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 ...
Yes, it means that considers decreasing sequences that converge to 0. I've only once worked with someone who preferred to use the
$↘$
and
$↗$
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
Prob. 15, Sec. 5.1, in Bartle & Sherbert's INTRO TO REAL ANALYSIS: A bounded function on
$(0,1)$
having no limit as
$x→0$
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 ...
What you did is correct. In order to show that
$α=β$
, suppose otherwise. That is, suppose that
$α=β$
. I will prove that
$lim_{x→0}f(x)=α(=β)$
, thereby reaching a ...
Use L'Hopital's with this problem?
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) ...
Let
$y=x→0_{+}lim (x1 )_{sinx},$
Now Let
$x=0+h,$
Then
$y=h→0lim (h1 )_{sinh}$
So
$ln(y)=h→0lim sin(h)⋅ln(h1 )=−h→0lim sinh⋅ln(h)=−h→0lim csc(h)ln(h) (∞∞ )$
...
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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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