1) A function is said to have a vertical asymptote wherever the limit on the left or right (or both) is either positive or negative infinity. For example, the function f(x)=(x^2 +1)/((2x+3)(x-5)) has a vertical asymptote at x=5
For each of the following limits, enter either 'P' for positive infinity, 'N' for negative infinity, or 'D' when the limit simply does not exist.
lim x→5 - x 2 +1(2x+3)(x-5) = limx→5-x2+1(2x+3)(x-5)=
lim x→5 + x 2 +1(2x+3)(x-5) = limx→5+x2+1(2x+3)(x-5)=
lim x→5 x 2 +1(2x+3)(x-5) = limx→5x2+1(2x+3)(x-5)=
2) Evaluate the following limits. If needed, enter INF for ∞ and MINF for -∞.
lim x→∞ (1+2x)/(3-5x) =
lim x→-∞ (1+2x)/(3-5x) =
3) Evaluate the following limits. If needed, enter INF for ∞ and MINF for -∞.
lim x→∞ (8x^3-6x^2)/(-9x10-2x-5x^3)
lim x→-∞ (8x^3-6x^2)/(-9x10-2x-5x^3)
4) Evaluate the following limits. If needed, enter INF for ∞ and MINF for -∞.
lim x→∞ (7x+88x^2)/(-5x+5)=
lim x→-∞ (7x+88x^2)/(-5x+5)=
5) Evaluate the following limits. If needed, enter INF for ∞and MINF for -∞.
limx→∞(√(3+3x^2))/(2+5x))=
limx→-∞(√(3+3x^2))/(2+5x))=
6) Enter I for ∞, -I for -∞, and DNE if the limit does not exist.
limx→∞(√(4x^2+x)-2x)
Limit =
7) Enter I for ∞, -I for -∞, and DNE if the limit does not exist.
limx→∞(√(x^2+5)-√(x^2-10))
Limit =
1) Note: Input inf for ∞∞, -inf for -∞-∞ or dne if needed.
limx→∞9cosx
Limit =
2) Evaluate the following limits. If needed, enter 'INF' for ∞ and '-INF' for -∞.
limx→∞(-33x^2-15x^3)=
limx→-∞(-33x^2-15x^3)=
3) Evaluate the following limit
limx→∞((9-√(x))/(9+√(x))
4) Evaluate the following limits. If needed, enter inf for ∞ and -inf -∞.
limx→∞x^2(-2+9x)(-9-3x)=
limx→-∞x^2(-2+9x)(-9-3x)=
5) Find the equations of the horizontal asymptotes and the vertical asymptotes of f(x)f(x). If there are no asymptotes of a given type, enter NONE. If there is more than one asymptote of a given type, give a comma separated list (i.e.: 1, 2,...).
f(x)=((x^2+x-6)/(3x^2+7x-6))
Horizontal asymptotes: y=
Vertical Asymptotes: x=
6) Evaluate the following limits,use "infinity" for "∞" and "-infinity" for "-∞".
limx→5-((2)/(x-5)^3)) =
limx→3+((2)/(x-3))=
limx→3-((2)/(x-3))=
limx→-7-((1)/(x^2(x+7))=