Parhs
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 Feb3 asked Integral proof $I_n=\int\frac{x^n}{\sqrt{x^2+5}} \, dx$ Feb3 accepted Evaluate $\int\sqrt[5]{\frac{x+5}{x-5}}\,\mathrm dx$ Feb3 comment Taylor Series $(x+2)/(2-3x)$ at $x=2$ according to wolfram alpha it seems that $3^n$ should be $3^{n+1}$ Feb3 comment Taylor Series $(x+2)/(2-3x)$ at $x=2$ to those who want to close this why?. Feb3 asked Taylor Series $(x+2)/(2-3x)$ at $x=2$ Feb3 asked Taylor series $\ln(x+3)$ at $x=1$ Feb2 comment Find monotony of $f(x)=x^5 +x^4 -11x^3 + 9x^2$ @namsap because it isnt easy to find roots for derivative Feb1 asked Find monotony of $f(x)=x^5 +x^4 -11x^3 + 9x^2$ Feb1 comment How to find $\lim_{x\to\infty}{\frac{e^x}{x^a}}$? Why is it correct to take log ? Jan27 comment Root with bolzano theorem cause theory has < . it makes sense but formally not Jan27 asked Root with bolzano theorem Jan27 accepted Limit find $f(x)$ $\lim_{x\to1}\frac{f(x)\cos{\frac{\pi x}{2}}}{\sqrt[3]x-1}=3$ Jan27 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... Jan27 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... Jan27 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 ? Jan27 asked Limit find $f(x)$ $\lim_{x\to1}\frac{f(x)\cos{\frac{\pi x}{2}}}{\sqrt[3]x-1}=3$ Jan27 accepted Series proof $\sum_1^\infty|a_n|<\infty$ then show that $\sum_1^\infty{a_n^2}<\infty$ Jan27 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 . Jan26 comment Limit evaluation $\lim_{x\to2}\frac{\sqrt{12-x^3}-\sqrt[3]{x^2+4}}{x^2-4}$ seems possible but hard Jan26 comment Limit evaluation $\lim_{x\to2}\frac{\sqrt{12-x^3}-\sqrt[3]{x^2+4}}{x^2-4}$ very smart... wow