Reputation
2,388
Top tag
Next privilege 2,500 Rep.
Create tag synonyms
Badges
3 15 35
Impact
~179k people reached

Apr
4
awarded  Popular Question
Apr
1
awarded  Nice Question
Jan
13
awarded  Yearling
Dec
14
asked What is the exact role of the integrals in a PID Controller?
Dec
8
accepted Solving $\int \left ( \frac{5x^2+3x-2}{x^3+2x^2} \right )$ via partial fractions
Dec
8
asked Solving $\int \left ( \frac{5x^2+3x-2}{x^3+2x^2} \right )$ via partial fractions
Dec
6
accepted Solving $\int \frac{1}{2x^2+3x+2}dx$ with partial fractions
Dec
6
revised Solving $\int \frac{1}{2x^2+3x+2}dx$ with partial fractions
edited body
Dec
6
asked Solving $\int \frac{1}{2x^2+3x+2}dx$ with partial fractions
Dec
5
awarded  Popular Question
Nov
15
asked How can I calculate the coordinates of the two extremes of a line when it is rotated?
Nov
14
awarded  Popular Question
Nov
10
asked Help understanding oblique asymptotes for $f(x) = 2x^3+3x^2-12x$
Nov
10
accepted Using L'Hopital to solve $\lim_{x\to +\infty}\frac{\frac{-1}{x^2}}{\sin^2\left(\frac{2}{x}\right)}$
Nov
10
comment Using L'Hopital to solve $\lim_{x\to +\infty}\frac{\frac{-1}{x^2}}{\sin^2\left(\frac{2}{x}\right)}$
Why did you pick $\frac{2}{x}$ in particular? I mean, I can see why your solution works, but I'm having a hard time understanding your reasoning behind picking $\frac{2}{x}$ and not something else. Is it because of the sine function?
Nov
10
comment Using L'Hopital to solve $\lim_{x\to +\infty}\frac{\frac{-1}{x^2}}{\sin^2\left(\frac{2}{x}\right)}$
I can see and understand why this works. However, it's not clear to me what triggered your idea of dividing $\frac{-1}{x^2}$ by $\frac{-1}{4}$. What made you see that?
Nov
10
asked Using L'Hopital to solve $\lim_{x\to +\infty}\frac{\frac{-1}{x^2}}{\sin^2\left(\frac{2}{x}\right)}$
Nov
9
accepted Solving $\lim_{x\to-\infty}x^2\cdot e^x$ with L'Hopital
Nov
9
comment Solving $\lim_{x\to-\infty}x^2\cdot e^x$ with L'Hopital
Do we have to change $x\to -\infty$ to $x \to \infty$? Can't we just keep $e^{-x}$? Because when you evaluate, it will become $e^{--\infty}$ anyway.
Nov
9
asked Solving $\lim_{x\to-\infty}x^2\cdot e^x$ with L'Hopital