Advertisements
Advertisements
Question
Evaluate the following :
`lim_(x -> 1) [(x + 3x^2 + 5x^3 + ... + (2"n" - 1)x^"n" - "n"^2)/(x - 1)]`
Advertisements
Solution
`lim_(x -> 1) [(x + 3x^2 + 5x^3 + ... + (2"n" - 1)x^"n" - "n"^2)/(x - 1)]`
1 + 3 + 5 + … + (2n – 1)
= `sum_("r" = 1)^"n" (2"r" - 1)`
= `2 sum_("r" = 1)^"n" "r" - sum_("r" = 1)^"n" 1`
= `2("n"("n" + 1))/2 - "n"`
= n(n + 1) – n
= n2 + n – n
= n2
∴ n2 = 1 + 3 + 5 + … + (2n – 1).
∴ Required limit
= `lim_(x -> 1) ([ x + 3x^2 + 5x^3 + ... + (2"n" - 1)x^"n"] - [1 + 3 + 5 + ... + (2"n" - 1)])/(x - 1)`
= `lim_(x -> 1) ((x - 1) + (3x^2 - 3) + (5x^2 - 5) + ... + (2"n" - 1)x^"n" - (2"n" - 1))/(x - 1)`
= `lim_(x -> 1) [(x - 1)/(x - 1) + (3(x^2 - 1))/(x - 1) + (5(x^3 - 1))/(x - 1) + ... + ((2"n" - 1)(x^"n" - 1))/(x - 1)]`
= `lim_(x -> 1) ((x^1 - 1^1)/(x - 1)) + 3lim_(x -> 1) ((x^2 - 1^2)/(x - 1)) + 5lim_(x -> 1)((x^3 - 1^3)/(x - 1)) + ... + (2"n" - 1) lim_(x -> 1) ((x^"n" - 1^"n")/(x - 1))`
= 1(1)0 + 3(2)(1)1 + 5(3)(1)2 + … + (2n – 1) n(1)n–1
= 1(1) + 3(2) + 5(3) + … + (2n – 1)n
= `sum_("r" = 1)^"n" (2"r" - 1)"r"`
= `sum_("r" = 1)^"n" (2"r"^2 - "r")`
= `2 sum_("r" = 1)^"n" "r"^2 - sum_("r" = 1)^"n" "r"`
= `2* ("n"("n" + 1)(2"n" + 1))/6 - ("n"("n" + 1))/2`
= `"n"("n" + 1)((2"n" + 1)/3 - 1/2)`
= `"n"("n" + 1) ((4"n" + 2- 3)/6)`
= `("n"("n" + 1) (4"n" - 1))/6`
APPEARS IN
RELATED QUESTIONS
Evaluate the following limit:
`lim_(x -> 3)[sqrt(2x + 6)/x]`
Evaluate the following limit:
`lim_(x -> 2)[(x^(-3) - 2^(-3))/(x - 2)]`
Evaluate the following limit:
`lim_(x -> 5)[(x^3 - 125)/(x^5 - 3125)]`
Evaluate the following limit :
`lim_(x -> 7) [(x^3 - 343)/(sqrt(x) - sqrt(7))]`
In problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> 0) (sqrt(x + 3) - sqrt(3))/x`
| x | – 0.1 | – 0.01 | – 0.001 | 0.001 | 0.01 | 0.1 |
| f(x) | 0.2911 | 0.2891 | 0.2886 | 0.2886 | 0.2885 | 0.28631 |
In problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> - 3) (sqrt(1 - x) - 2)/(x + 3)`
| x | – 3.1 | – 3.01 | – 3.00 | – 2.999 | – 2.99 | – 2.9 |
| f(x) | – 0.24845 | – 0.24984 | – 0.24998 | – 0.25001 | – 0.25015 | – 0.25158 |
In problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> 0) sin x/x`
| x | – 0.1 | – 0.01 | – 0.001 | 0.001 | 0.01 | 0.1 |
| f(x) | 0.99833 | 0.99998 | 0.99999 | 0.99999 | 0.99998 | 0.99833 |
In exercise problems 7 – 15, use the graph to find the limits (if it exists). If the limit does not exist, explain why?
`lim_(x -> 1) (x^2 + 2)`
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.
If f(2) = 4, can you conclude anything about the limit of f(x) as x approaches 2?
Evaluate the following limits:
`lim_("h" -> 0) (sqrt(x + "h") - sqrt(x))/"h", x > 0`
Evaluate the following limits:
`lim_(x -> 5) (sqrt(x + 4) - 3)/(x - 5)`
Evaluate the following limits:
`lim_(x -> 0) (sqrt(1 + x) - 1)/x`
Evaluate the following limits:
`lim_(x -> 2) (2 - sqrt(x + 2))/(root(3)(2) - root(3)(4 - x))`
Evaluate the following limits:
`lim_(x - 0) (sqrt(1 + x^2) - 1)/x`
Find the left and right limits of f(x) = tan x at x = `pi/2`
Evaluate the following limits:
`lim_(x -> 3) (x^2 - 9)/(x^2(x^2 - 6x + 9))`
Show that `lim_("n" -> oo) 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/("n"("n" + 1))` = 1
A tank contains 5000 litres of pure water. Brine (very salty water) that contains 30 grams of salt per litre of water is pumped into the tank at a rate of 25 litres per minute. The concentration of salt water after t minutes (in grams per litre) is C(t) = `(30"t")/(200 + "t")`. What happens to the concentration as t → ∞?
Evaluate the following limits:
`lim_(x -> oo) ((2x^2 + 3)/(2x^2 + 5))^(8x^2 + 3)`
Evaluate the following limits:
`lim_(x -> 0) (2 "arc"sinx)/(3x)`
Evaluate the following limits:
`lim_(x-> 0) (1 - cos x)/x^2`
Evaluate the following limits:
`lim_(x -> 0) (tan 2x)/x`
Evaluate the following limits:
`lim_(x -> 0) (1 - cos^2x)/(x sin2x)`
Evaluate the following limits:
`lim_(x - oo){x[log(x + "a") - log(x)]}`
Evaluate the following limits:
`lim_(x -> pi) (sin3x)/(sin2x)`
Evaluate the following limits:
`lim_(x -> pi) (1 + sinx)^(2"cosec"x)`
Evaluate the following limits:
`lim_(x -> 0) (sqrt(2) - sqrt(1 + cosx))/(sin^2x)`
Evaluate the following limits:
`lim_(x -> 0) (sqrt(1 + sinx) - sqrt(1 - sinx))/tanx`
Evaluate the following limits:
`lim_(x -> ) (sinx(1 - cosx))/x^3`
Evaluate the following limits:
`lim_(x -> 0) (tan x - sin x)/x^3`
Choose the correct alternative:
`lim_(x -> 0) sqrt(1 - cos 2x)/x`
Choose the correct alternative:
If `lim_(x -> 0) (sin "p"x)/(tan 3x)` = 4, then the value of p is
Choose the correct alternative:
`lim_(x -> 0) ("e"^tanx - "e"^x)/(tan x - x)` =
If `lim_(x -> 1) (x + x^2 + x^3|+ .... + x^n - n)/(x - 1)` = 820, (n ∈ N) then the value of n is equal to ______.
`lim_(x→∞)((x + 7)/(x + 2))^(x + 4)` is ______.
