Advertisements
Advertisements
प्रश्न
Show that `lim_("n" -> oo) 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/("n"("n" + 1))` = 1
Advertisements
उत्तर
Tn = `1/("n"("n" + 1))`
`1/("n"("n" + 1)) = "A"/"n" + "B"/("n" +1)`
1 = A(n + 1) + Bn
Put n = 0
1 = A(0 + 1) + B × 0
A = 1
Put n = – 1
1 = A(– 1 + 1) + B(– 1)
1 = 0 – B
⇒ B = – 1
`1/("n"("n" + 1)) = 1/"n" - 1/("n" + 1)`
Tn = `1/"n" - 1/("n" + 1)`
T1 = `1/1- 1/2`
T2 = `1/2 - 1/3`
T3 = `1/3 - 1/4`
........................
........................
Tn = `1/"n" - 1/("n" + 1)`
T1 + T2 + .... + Tn = `(1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + .... + (1/"n" - 1/("n" + 1))`
T1 + T2 + .... + Tn = `(1 - 1/("n" + 1))`
`lim_("n" -> oo) ("T"_1 + "T"_2 + .... + "T"_"n") = lim_("n" -> oo) (1 - 1/("n" + 1))`
= `1 - lim_("n" -> oo) 1/("n" + 1)`
= 1 – 0
`lim_("n" -> oo) 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/("n"("n" + 1))` = 1
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit:
`lim_(x -> 5)[(x^3 - 125)/(x^5 - 3125)]`
Evaluate the following :
`lim_(x -> 0)[x/(|x| + x^2)]`
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 |
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 |
Sketch the graph of f, then identify the values of x0 for which `lim_(x -> x_0)` f(x) exists.
f(x) = `{{:(x^2",", x ≤ 2),(8 - 2x",", 2 < x < 4),(4",", x ≥ 4):}`
Verify the existence of `lim_(x -> 1) f(x)`, where `f(x) = {{:((|x - 1|)/(x - 1)",", "for" x ≠ 1),(0",", "for" x = 1):}`
Evaluate the following limits:
`lim_(x -> 2) (1/x - 1/2)/(x - 2)`
Evaluate the following limits:
`lim_(x -> 0) (sqrt(x^2 + 1) - 1)/(sqrt(x^2 + 16) - 4)`
Evaluate the following limits:
`lim_(x -> 1) (root(3)(7 + x^3) - sqrt(3 + x^2))/(x - 1)`
Evaluate the following limits:
`lim_(x -> oo) (x^4 - 5x)/(x^2 - 3x + 1)`
Show that `lim_("n" -> oo) (1^2 + 2^2 + ... + (3"n")^2)/((1 + 2 + ... + 5"n")(2"n" + 3)) = 9/25`
Evaluate the following limits:
`lim_(alpha -> 0) (sin(alpha^"n"))/(sin alpha)^"m"`
Evaluate the following limits:
`lim_(x -> 0) (tan 2x)/x`
Evaluate the following limits:
`lim_(x -> pi) (1 + sinx)^(2"cosec"x)`
Evaluate the following limits:
`lim_(x -> 0) (tan x - sin x)/x^3`
Choose the correct alternative:
`lim_(x - pi/2) (2x - pi)/cos x`
Choose the correct alternative:
If `lim_(x -> 0) (sin "p"x)/(tan 3x)` = 4, then the value of p is
`lim_(x→0^+)(int_0^(x^2)(sinsqrt("t"))"dt")/x^3` is equal to ______.
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 ______.
