Advertisements
Advertisements
प्रश्न
Show that `lim_("n" -> oo) (1^2 + 2^2 + ... + (3"n")^2)/((1 + 2 + ... + 5"n")(2"n" + 3)) = 9/25`
Advertisements
उत्तर
`lim_("n" -> oo) (1^2 + 2^2 + ... + (3"n")^2)/((1 + 2 + ... + 5"n")(2"n" + 3)) = lim_("n" -> oo) ((3"n"(3"n" + 1)(2 xx 3"n" + 1))/6)/((5"n"(5"n" + 1))/2 (2"n" + 3))`
= `lim_("n" -> oo) (3"n"(3"n" + 1)(6"n" + 1) xx 2)/(6 xx 5"n"(5"n" + 1)(2"n" + 3))`
= `lim_("n" -> oo) ("n"(3"n" + 1)(6"n" + 1))/(5"n"(5"n" + 1)(2"n" + 3))`
= `lim_("n" -> oo) ("n"*"n"(3 + 1/"n") "n"(6 + 1/"n"))/(5"n"*"n"(5 + 1/"n") "n"*(2 + 3/"n"))`
= `lim_("n" -> oo) ("n"^3(3 + 1/"n")(6 + 1/"n"))/("n"^3*5(5 + 1/"n")(2 + 3/"n"))`
= `((3 + 0)(6 + 0))/(5(5 + 0)(2 + 0))`
= `18/50`
= `9/25`
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit:
`lim_(z -> -3) [sqrt("z" + 6)/"z"]`
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 exercise problems 7 – 15, use the graph to find the limits (if it exists). If the limit does not exist, explain why?
`lim_(x -> 0) sec x`
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):}`
Sketch the graph of f, then identify the values of x0 for which `lim_(x -> x_0)` f(x) exists.
f(x) = `{{:(sin x",", x < 0),(1 - cos x",", 0 ≤ x ≤ pi),(cos x",", x > pi):}`
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 -> 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 -> 5) (sqrt(x - 1) - 2)/(x - 5)`
Evaluate the following limits:
`lim_(x -> "a") (sqrt(x - "b") - sqrt("a" - "b"))/(x^2 - "a"^2) ("a" > "b")`
Evaluate the following limits:
`lim_(x -> oo) 3/(x - 2) - (2x + 11)/(x^2 + x - 6)`
Evaluate the following limits:
`lim_(x -> 0) (2^x - 3^x)/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) (1 + sinx)^(2"cosec"x)`
Evaluate the following limits:
`lim_(x -> 0) ("e"^x - "e"^(-x))/sinx`
Choose the correct alternative:
`lim_(x -> oo) ((x^2 + 5x + 3)/(x^2 + x + 3))^x` is
Choose the correct alternative:
`lim_(x - oo) sqrt(x^2 - 1)/(2x + 1)` =
Choose the correct alternative:
`lim_(x -> oo) (1/"n"^2 + 2/"n"^2 + 3/"n"^2 + ... + "n"/"n"^2)` is
`lim_(x -> 0) ((2 + x)^5 - 2)/((2 + x)^3 - 2)` = ______.
