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Asymptote of a curve in polar coordinates - Physics Forums The curve C has polar equation r θ = 1 for 0 <θ <2 π Use the fact that lim θ → 0 s i n θ θ = 1 to show the line y = 1 is an asymptote to C The Attempt at a Solution **Attempt** I understood the concept behind how this asymptote is calculated, but I am not very fluent in mathematics to convert the above information into a comprehensive
Determining the horizontal asymptote - Physics Forums I am self-studying; My interest is on the horizontal asymptote, now considering the degree of polynomial and leading coefficients, i have Therefore is the horizontal asymptote The part that i do not seem to get is (i already checked this on desmos) why an asymptote can be regarded as such if it is crossing the curve In my small understanding, i thought asymptote ought not to intersect any
Oblique Asymptotes: What happens to the Remainder? An "asymptote" is a line that a curve approaches as x goes to, in this case, negative infinity and infinity Yes, long division gives a quotient of -3x- 3 with a remaider of -1 That means that As x goes to either infinity or negative infinity, that last fraction goes to 0
Horizontal asymptotes - approaches from above or below? I seem to be having a lot of difficulty finding whether for a horizontal asymptote, whether the curve approaches the asymptote from above or below For example, for the problem y = \\frac{6x + 1}{1 - 2x}, I know that: For the vertical asymptote, x = 1 2, and that \\lim_{x \\to
Describing behavior on each side of a vertical asymptote Find the vertical asymptotes of the graph of F (x) = (3 - x) (x^2 - 16) ok if i factor the denominator i find the vertical asymptotes to be x = 4, x = -4 The 2nd part of the problem asks: Describe the behavior of f (x) to the left and right of each vertical asymptote I'm not sure what i need to write for this