Advertisements
Advertisements
प्रश्न
Evaluate the following :
`lim_(x -> 1) [(sqrt(x) - 1)/logx]`
उत्तर
`lim_(x -> 1) (sqrt(x) - 1)/logx`
put x – 1 = h
∴ x = 1 + h
As x → 1, h → 0
∴ Required limit
= `lim_("h" -> 0) (sqrt(1 + "h") - 1)/(log(1 + "h"))`
= `lim_("h" -> 0) (sqrt(1 + "h") - 1)/(log (1 + "h")) xx (sqrt(1 + "h") + 1)/(sqrt(1 + "h") + 1)`
= `lim_("h" -> 0) ((1 + "h") - 1)/(log(1 + "h")) xx 1/(sqrt(1 + "h") + 1)`
= `lim_("h" -> 0) "h"/(log(1 + "h")) xx 1/(sqrt(1 + "h") + 1)`
= `lim_("h" -> 0) (1)/((log(1 + "h"))/("h")) xx 1/(sqrt(1 + "h") + 1) ...[("Divide Numerator and"),("Denominator by h"),("As" "h" -> 0"," "h" ≠ 0)]`
= `lim_("h" -> 0) (1)/((log(1 + "h"))/("h")) xx 1/(lim_("h" -> 0)(sqrt(1 + "h") + 1)`
= `1/1 xx 1/(sqrt(1 + 0) + 1)`
= `1/2`
APPEARS IN
संबंधित प्रश्न
Evaluate the following :
`lim_(x -> pi/2) [("cosec"x - 1)/(pi/2 - x)^2]`
Evaluate the following :
`lim_(x -> "a") [(sinx - sin"a")/(root(5)(x) - root(5)("a"))]`
Evaluate the following :
`lim_(x -> pi) [(sqrt(5 + cosx) - 2)/(pi - x)^2]`
Evaluate the following :
`lim_(x -> pi/6) [(cos x - sqrt(3) sinx)/(pi - 6x)]`
Evaluate the following :
`lim_(x -> 1) [(1 - x^2)/(sinpix)]`
Evaluate the following:
`lim_(x→π/6) [(2sinx − 1)/(π − 6x)]`
Evaluate the following :
`lim_(x -> pi/4) [(sqrt(2) - cosx - sinx)/(4x - pi)^2]`
Evaluate the following :
`lim_(x -> pi/6) [(2 - sqrt(3)cosx - sinx)/(6x - pi)^2]`
Evaluate the following :
`lim_(x -> "a") [(sin(sqrt(x)) - sin(sqrt("a")))/(x - "a")]`
Evaluate the following:
`lim_(x -> pi/2) [(cos3x + 3cosx)/(2x - pi)^3]`
Evaluate the following :
`lim_(x -> 0)[((1 - x)^5 - 1)/((1 - x)^3 - 1)]`
Evaluate the following :
`lim_(x -> 1) [(2^(2x - 2) - 2^x + 1)/(sin^2 (x - 1))]`
Evaluate the following :
`lim_(x -> 1) [(4^(x - 1) - 2^x + 1)/(x - 1)^2]`
`lim_{n→∞}[1/(n^2 + 1) + 2/(n^2 + 1) + 3/(n^2 + 1) + .... + n/(n^2 + 1)]` = ______
