Advertisements
Advertisements
Question
`lim_(y -> 0) ((x + y) sec(x + y) - x sec x)/y`
Advertisements
Solution
`lim_(y -> 0) ((x + y) sec(x + y) - x sec x)/y`
= `lim_(y -> 0) (x sec(x + y) + y sec (x + y) - x sec x)/y`
= `lim_(y -> 0) ([x sec (x + y) - x sec x])/y + lim_(y -> 0) (y sec (x + y))/y`
= `lim_(y -> 0) (x[sec(x + y) - sec x])/y + lim_(y -> 0) sec (x + y)`
= `lim_(y -> 0) (x[1/(cos(x + y)) - 1/cosx])/y + lim_(y -> 0) sec(x + y)`
= `lim_(y -> 0) x[(cosx - cos(x + y))/(y * cos(x + y) * cos x)] + lim_(y -> 0) sec(x + y)`
= `lim_(y -> 0) (x[-2 sin ((x + x + y)/2) * sin ((x - x - y)/2)])/(y cos(x + y) * cos x) + lim_(y -> 0) sec(x + y)`
= `(x[- 2 sin (x + y/2) * sin(- y/2)])/(cos(x + y) * cos x * y) + lim_(y -> 0) sec(x + y)`
= `lim_((y -> 0),(because y/2 -> 0)) x[([2 sin (x + y/2) sin (y/2)])/(cos (x + y) * cos x * (y/2) * 2)] + lim_(y -> 0) sec(x + y)`
∴ Taking the limits we have
= `x[sin x * 1/(cosx * cos x)] + sec x`
= `x sec x tan x + sec x`
= `sec x(x tan x + 1)`
APPEARS IN
RELATED QUESTIONS
Evaluate the following limit.
`lim_(x -> pi) (sin(pi - x))/(pi (pi - x))`
Evaluate the following limit.
`lim_(x -> 0) (cos 2x -1)/(cos x - 1)`
Evaluate the following limit.
`lim_(x -> (pi)/2) (tan 2x)/(x - pi/2)`
Evaluate the following limit :
`lim_(theta -> 0) [(1 - cos2theta)/theta^2]`
Evaluate the following limit :
`lim_(x -> 0) [(x*tanx)/(1 - cosx)]`
Evaluate the following limit :
`lim_(x ->0)((secx - 1)/x^2)`
Evaluate the following limit :
`lim_(x -> pi) [(sqrt(1 - cosx) - sqrt(2))/(sin^2 x)]`
Evaluate the following limit :
`lim_(x -> pi/4) [(tan^2x - cot^2x)/(secx - "cosec"x)]`
Evaluate the following limit :
`lim_(x -> pi/6) [(2sin^2x + sinx - 1)/(2sin^2x - 3sinx + 1)]`
Evaluate the following :
`lim_(x -> 0)[(secx^2 - 1)/x^4]`
Evaluate the following :
`lim_(x -> pi/4) [(sinx - cosx)^2/(sqrt(2) - sinx - cosx)]`
`lim_{x→0}((3^x - 3^xcosx + cosx - 1)/(x^3))` is equal to ______
Evaluate `lim_(x -> 0) (sqrt(2 + x) - sqrt(2))/x`
Evaluate `lim_(x -> 0) (tanx - sinx)/(sin^3x)`
Evaluate `lim_(x -> a) (sqrt(a + 2x) - sqrt(3x))/(sqrt(3a + x) - 2sqrt(x))`
Find the derivative of f(x) = `sqrt(sinx)`, by first principle.
`lim_(x -> 0) |x|/x` is equal to ______.
Evaluate: `lim_(x -> 3) (x^2 - 9)/(x - 3)`
Evaluate: `lim_(x -> pi/3) (sqrt(1 - cos 6x))/(sqrt(2)(pi/3 - x))`
Evaluate: `lim_(x -> pi/6) (sqrt(3) sin x - cos x)/(x - pi/6)`
Evaluate: `lim_(x -> 0) (sin 2x + 3x)/(2x + tan 3x)`
Evaluate: `lim_(x -> 0) (sqrt(2) - sqrt(1 + cos x))/(sin^2x)`
Evaluate: `lim_(x -> 0) (sin x - 2 sin 3x + sin 5x)/x`
cos (x2 + 1)
x cos x
`lim_(x -> 0) (x^2 cosx)/(1 - cosx)` is ______.
`lim_(x -> 0) ((1 + x)^n - 1)/x` is equal to ______.
`lim_(x -> 0) (1 - cos 4theta)/(1 - cos 6theta)` is ______.
`lim_(x -> 0) ("cosec" x - cot x)/x` is equal to ______.
`lim_(x -> 0) sinx/(sqrt(x + 1) - sqrt(1 - x)` is ______.
If `f(x) = {{:(sin[x]/[x]",", [x] ≠ 0),(0",", [x] = 0):}`, where [.] denotes the greatest integer function, then `lim_(x -> 0) f(x)` is equal to ______.
`lim_(x -> 3^+) x/([x])` = ______.
The value of `lim_(x → ∞) ((x^2 - 1)sin^2(πx))/(x^4 - 2x^3 + 2x - 1)` is equal to ______.
If L = `lim_(x→∞)(x^2sin 1/x - x)/(1 - |x|)`, then value of L is ______.
`lim_(x rightarrow π/2) ([1 - tan (x/2)] (1 - sin x))/([1 + tan (x/2)] (π - 2x)^3` is ______.
`lim_(theta → -pi/4) (cos theta + sin theta)/(theta + pi/4)` =
