/sci/ - Science & Math

>>14959013
>>14959013
>>14959022
Feeding you a spoonful
>I don't know anything about arc tan,arc cos, etc
https://en.wikipedia.org/wiki/Differentiation_of_trigonometric_functions

>(a)
[eqn] \lim_{x\rightarrow0}\frac{x}{\tan^{-1}3x}=\lim_{x\rightarrow0}\frac{\frac{d}{dx}\left(x\right)}{\frac{d}{dx}\left(\tan^{-1}3x\right)}=\lim_{x\rightarrow0}\frac{1}{\frac{3}{\tan^{2}(\tan^{-1}(3x))+1}}=\lim_{x\rightarrow0}\frac{1}{\frac{3}{9x^{2}+1}}=\lim_{x\rightarrow0}\frac{9x^{2}+1}{3}=\frac{1}{3} [/eqn]

>(b)
Literally the same example from wikipedia but multiplied by 3, so the answer should be 3/2
https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule#Other_indeterminate_forms

>(c)
[eqn] \lim_{x\rightarrow0}\frac{\cos mx-\cos nx}{x^{2}}=\lim_{x\rightarrow0}\frac{-m\sin mx+n\sin nx}{2x}=\lim_{x\rightarrow0}\frac{-m^{2}\cos mx+n^{2}\cos nx}{2}=\frac{-m^{2}+n^{2}}{2} [/eqn]

>(d)
[eqn] \text{Let }L=\lim_{x\rightarrow\infty}\left(1+\frac{a}{x}\right)^{bx}\\\ln L=\ln\left[\lim_{x\rightarrow\infty}\left(1+\frac{a}{x}\right)^{bx}\right]=\lim_{x\rightarrow\infty}\ln\left[\left(1+\frac{a}{x}\right)^{bx}\right]=\lim_{x\rightarrow\infty}bx\ln\left(1+\frac{a}{x}\right)=\lim_{x\rightarrow\infty}b\frac{\ln\left(1+\frac{a}{x}\right)}{x^{-1}}=\lim_{x\rightarrow\infty}b\frac{-\left(1+\frac{a}{x}\right)^{-1}ax^{-2}}{-x^{-2}}=\lim_{x\rightarrow\infty}b\left(1+\frac{a}{x}\right)^{-1}a=ab\\\Rightarrow\lim_{x\rightarrow\infty}\left(1+\frac{a}{x}\right)^{bx}=L=e^{\ln L}=e^{ab} [/eqn]