Advertisements
Advertisements
Question
Evaluate the following limits:
`lim_(x ->oo) (x^3/(2x^2 - 1) - x^2/(2x + 1))`
Advertisements
Solution
`lim_(x ->oo) (x^3/(2x^2 - 1) - x^2/(2x + 1)) = lim_(x -> oo) [(x^2(2x + 1) - x^2(2x^2 - 1))/((2x^2 - 1)(2x + 1))]`
= `lim_(x -> oo) [(2x^4 + x^3 - 2x^4 +x^2)/((2x^2 - 1)(2x + 1))]`
= `lim_(x -> oo) [(x^3 + x^2)/((2x^2 - 1)(2x + 1))]`
= `lim_(x - oo) [(x^3 (1 + x^2/x^3))/(x^2(2 - 1/x^2) xx (2 + 1/x))]`
= `lim_(x - oo) [((1 + 1/x))/((2 - 1/x^2) (2 + 1/x))]`
= `((1 + 1/oo))/((2 - 1/oo) (2 + 1/oo))`
= `(1 + 0)/((2 - 0) (2 + 0))`
`lim_(x ->oo) (x^3/(2x^2 - 1) - x^2/(2x + 1)) = 1/4`
APPEARS IN
RELATED QUESTIONS
Evaluate the following limit:
`lim_(y -> -3) [(y^5 + 243)/(y^3 + 27)]`
Evaluate the following limit :
`lim_(x -> 0)[((1 - x)^8 - 1)/((1 - x)^2 - 1)]`
In the following example, given ∈ > 0, find a δ > 0 such that whenever, |x – a| < δ, we must have |f(x) – l| < ∈.
`lim_(x -> 2)(2x + 3)` = 7
In the following example, given ∈ > 0, find a δ > 0 such that whenever, |x – a| < δ, we must have |f(x) – l| < ∈.
`lim_(x -> 1) (x^2 + x + 1)` = 3
Evaluate the following :
Given that 7x ≤ f(x) ≤ 3x2 – 6 for all x. Determine the value of `lim_(x -> 3) "f"(x)`
In problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> 2) (x - 2)/(x^2 - 4)`
| x | 1.9 | 1.99 | 1.999 | 2.001 | 2.01 | 2.1 |
| f(x) | 0.25641 | 0.25062 | 0.250062 | 0.24993 | 0.24937 | 0.24390 |
Sketch the graph of a function f that satisfies the given value:
f(– 2) = 0
f(2) = 0
`lim_(x -> 2) f(x)` = 0
`lim_(x -> 2) f(x)` does not exist.
Write a brief description of the meaning of the notation `lim_(x -> 8) f(x)` = 25
Evaluate the following limits:
`lim_(sqrt(x) -> 3) (x^2 - 81)/(sqrt(x) - 3)`
Evaluate the following limits:
`lim_(x -> 1) (sqrt(x) - x^2)/(1 - sqrt(x))`
Evaluate the following limits:
`lim_(x -> 0) (sqrt(x^2 + 1) - 1)/(sqrt(x^2 + 16) - 4)`
Evaluate the following limits:
`lim_(x -> 0) (sqrt(1 + x) - 1)/x`
Evaluate the following limits:
`lim_(x - 0) (sqrt(1 + x^2) - 1)/x`
Evaluate the following limits:
`lim_(x -> 0) (sqrt(1 - x) - 1)/x^2`
Evaluate the following limits:
`lim_(x -> 3) (x^2 - 9)/(x^2(x^2 - 6x + 9))`
Evaluate the following limits:
`lim_(x -> 0) (2 "arc"sinx)/(3x)`
Evaluate the following limits:
`lim_(x -> 0) ("e"^("a"x) - "e"^("b"x))/x`
Choose the correct alternative:
`lim_(x - pi/2) (2x - pi)/cos x`
