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Feb
3
asked Integral proof $I_n=\int\frac{x^n}{\sqrt{x^2+5}} \, dx$
Feb
3
accepted Evaluate $\int\sqrt[5]{\frac{x+5}{x-5}}\,\mathrm dx$
Feb
3
comment Taylor Series $(x+2)/(2-3x)$ at $x=2$
according to wolfram alpha it seems that $3^n$ should be $3^{n+1}$
Feb
3
comment Taylor Series $(x+2)/(2-3x)$ at $x=2$
to those who want to close this why?.
Feb
3
asked Taylor Series $(x+2)/(2-3x)$ at $x=2$
Feb
3
asked Taylor series $\ln(x+3)$ at $x=1$
Feb
2
comment Find monotony of $f(x)=x^5 +x^4 -11x^3 + 9x^2$
@namsap because it isnt easy to find roots for derivative
Feb
1
asked Find monotony of $f(x)=x^5 +x^4 -11x^3 + 9x^2$
Feb
1
comment How to find $\lim_{x\to\infty}{\frac{e^x}{x^a}}$?
Why is it correct to take log ?
Jan
27
comment Root with bolzano theorem
cause theory has < . it makes sense but formally not
Jan
27
asked Root with bolzano theorem
Jan
27
accepted Limit find $f(x)$ $ \lim_{x\to1}\frac{f(x)\cos{\frac{\pi x}{2}}}{\sqrt[3]x-1}=3 $
Jan
27
comment Limit find $f(x)$ $ \lim_{x\to1}\frac{f(x)\cos{\frac{\pi x}{2}}}{\sqrt[3]x-1}=3 $
Just solving some before LHospital problems and want to learn without it too...
Jan
27
comment Limit find $f(x)$ $ \lim_{x\to1}\frac{f(x)\cos{\frac{\pi x}{2}}}{\sqrt[3]x-1}=3 $
It seems the key was there to change cos to sin. I should have thought it...
Jan
27
comment Limit find $f(x)$ $ \lim_{x\to1}\frac{f(x)\cos{\frac{\pi x}{2}}}{\sqrt[3]x-1}=3 $
hm yep, however I have added without l'Hospital tag. Any idea there ?
Jan
27
asked Limit find $f(x)$ $ \lim_{x\to1}\frac{f(x)\cos{\frac{\pi x}{2}}}{\sqrt[3]x-1}=3 $
Jan
27
accepted Series proof $\sum_1^\infty|a_n|<\infty$ then show that $\sum_1^\infty{a_n^2}<\infty$
Jan
27
comment Series proof $\sum_1^\infty|a_n|<\infty$ then show that $\sum_1^\infty{a_n^2}<\infty$
I am going to ask why it is poor but probably he isnt going to tell me(cant insist he'll get annoyed).However after asking lots of other classmates someone showed me the solution . It is like this math.stackexchange.com/a/493782/17446 .
Jan
26
comment Limit evaluation $\lim_{x\to2}\frac{\sqrt{12-x^3}-\sqrt[3]{x^2+4}}{x^2-4}$
seems possible but hard
Jan
26
comment Limit evaluation $\lim_{x\to2}\frac{\sqrt{12-x^3}-\sqrt[3]{x^2+4}}{x^2-4}$
very smart... wow