This site uses cookies for analytics, personalized content and ads. By continuing to browse this site, you agree to this use.
Learn more
Microsoft Math Solver
Solve
Practice
Download
Solve
Practice
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
Algebra Calculator
Trigonometry Calculator
Calculus Calculator
Matrix Calculator
Download
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
Algebra Calculator
Trigonometry Calculator
Calculus Calculator
Matrix Calculator
Type a math problem
algebra
trigonometry
statistics
calculus
matrices
variables
list
AC
log
ln
(
)
7
8
9
τ
π
4
5
6
≤
≥
%
θ
1
2
3
<
>
x
i
0
.
y
Evaluate
5
$5$
Quiz
Limits
10 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
$ln(y)=h→0lim sin(h)⋅ln(h1 )=−h→0lim sinh⋅ln(h)=−h→0lim csc(h)ln(h) (∞∞ )$
...
More Items
Share
Copy
Copied to clipboard
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 $
Back to top
English
English
Deutsch
Español
Français
Italiano
Português
Русский
简体中文
繁體中文
Bahasa Melayu
Bahasa Indonesia
العربية
日本語
Türkçe
Polski
עברית
Čeština
Nederlands
Magyar Nyelv
한국어
Slovenčina
ไทย
ελληνικά
Română
Tiếng Việt
हिन्दी
অসমীয়া
বাংলা
ગુજરાતી
ಕನ್ನಡ
कोंकणी
മലയാളം
मराठी
ଓଡ଼ିଆ
ਪੰਜਾਬੀ
தமிழ்
తెలుగు