Advertisements
Advertisements
प्रश्न
Evaluate the following limits:
`lim_(x -> 0) (sqrt(x^2 + "a"^2) - "a")/(sqrt(x^2 + "b"^2) - "b")`
Advertisements
उत्तर
We know `lim_(x -> "a") (x^"n" - "a"^"a")/(x -"a") = "na"^("n" - 1)`
`lim_(x -> 0)((sqrt((x^2 + "a"^2)) - "a")/(sqrt((x^2 + "b"^2)) - "b")) = lim_(x -> 0)(((x^2 + "a"^2)^(1/2) - ("a"^2)^(1/2))/((x^2 + "b"^2)^(1/2) - ("b"^2)^(1/2)))`
= `lim_(x -> 0) ((x^2 + "a"^2)^(1/2) - ("a"^2)^(1/2))/x^2 xx x^2/((x^2 + "b"^2)^(1/2) - ("b"^2)^(1/2))`
= `lim_(x -> 0) ((x^2 + "a"^2)^(1/2) - ("a"^2)^(1/2))/((x^2 + "a"^2) - "a"^2) xx ((x^2 + "b"^2) - "b"^2)/((x^2 + "b"^2)^(1/2) - ("b"^2)^(1/2))`
= `lim_(x -> 0) ((x^2 + "a"^2)^(1/2) - ("a"^2)^(1/2))/((x^2 + "a"^2) - "a"^2) xx lim_(x -> 0) 1/(((x^2 + "b"^2)^(1/2) - ("b"^2)^(1/2))/((x^2 + "b"^2) - "b"^2)`
Put x2 + a2 = y
Put x2 + b2 = z
When x = 0
⇒ y = a2
When x = 0
⇒ z = b2
= `lim_(y -> "a"^2)((y^(1/2) - ("a"^2)^(1/))/(y - "a"^2)) xx 1/(lim_(x -> "b"^2)(("z"^(1/2) - ("b"^2)^(1/2))/("z" - "b"^2))`
= `1/2 ("a"^2)^(1/2 - 1) xx 1/(1/2 ("b"^2)^(1/2 - 1))`
= `(("a"^2)^(- 1/2))/(("b"^2)^(- 1/2))`
= `("a"^(- 1))/("b"^(- 1))`
= `"b"/"a"`
`lim_(x -> 0)((sqrt((x^2 + "a"^2)) - "a")/(sqrt((x^2 + "b"^2)) - "b")) = "b"/"a"`
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit:
`lim_(y -> -3) [(y^5 + 243)/(y^3 + 27)]`
Evaluate the following limit :
`lim_(x -> 1)[(x + x^2 + x^3 + ......... + x^"n" - "n")/(x - 1)]`
Evaluate the following limit :
`lim_(z -> "a")[((z + 2)^(3/2) - ("a" + 2)^(3/2))/(z - "a")]`
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 a function f that satisfies the given value:
f(0) is undefined
`lim_(x -> 0) f(x)` = 4
f(2) = 6
`lim_(x -> 2) f(x)` = 3
If f(2) = 4, can you conclude anything about the limit of f(x) as x approaches 2?
Evaluate the following limits:
`lim_(x -> 2) (x^4 - 16)/(x - 2)`
Evaluate the following limits:
`lim_("h" -> 0) (sqrt(x + "h") - sqrt(x))/"h", x > 0`
Evaluate the following limits:
`lim_(x -> 1) (root(3)(7 + x^3) - sqrt(3 + x^2))/(x - 1)`
Evaluate the following limits:
`lim_(x -> 2) (2 - sqrt(x + 2))/(root(3)(2) - root(3)(4 - x))`
Evaluate the following limits:
`lim_(x -> "a") (sqrt(x - "b") - sqrt("a" - "b"))/(x^2 - "a"^2) ("a" > "b")`
Show that `lim_("n" -> oo) (1^2 + 2^2 + ... + (3"n")^2)/((1 + 2 + ... + 5"n")(2"n" + 3)) = 9/25`
Show that `lim_("n" -> oo) 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/("n"("n" + 1))` = 1
Evaluate the following limits:
`lim_(alpha -> 0) (sin(alpha^"n"))/(sin alpha)^"m"`
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)]}`
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_(alpha - pi/4) (sin alpha - cos alpha)/(alpha - pi/4)` is
The value of `lim_(x→0)(sin(ℓn e^x))^2/((e^(tan^2x) - 1))` is ______.
