Oblique Asymptotes (3 of 3: Investigating shapes)

[@Analyst101]

I must say the way mr. Woo presents math is great. I wish I had a teacher like him.

[@izakj5094]

Josip Šajrih you do now :)

[@michaelempeigne3519]

to find a slant asymptote, i find using only limits works : a = lim ( x -- > inf ) [ f ( x ) / x ] = lim ( x -- > inf ) [ ( 1 - x^2 ) / x^2 ] = lim ( x -- > inf ) [ ( 1 / x^2 ) - 1 ] = lim ( x -- > inf ) [ ( 1 / x^2 ) ] - 1 = 0 - 1 = - 1 b = lim ( x -- > inf ) [ f ( x ) - ax ] = lim ( x -- > inf ) [ ( 1 - x^2 ) / x + x ] = lim ( x -- > inf ) [ ( 1 - x^2 + x^2 ) / x ] = lim ( x -- > inf ) [ 1 / x ] = 0 the slant asymptote is y = ax + b = - x + 0 = - x

[@yoavshati]

10:03 It's called a curvilinear asymptotes in general, or a parabolic asymptote in this case

[@andrewps339]

(1-x^2)/x is -x with a remainder. The remainder always approaches 0 so it’s not useful. Which means the oblique asymptote is -x. What was the point of the calculus. I would like to know. Thank you

[@021sunnykumar5]

sir graphically how we can say that integration of (cos x ) 0 to infinity is 1