English

Limy→0(x+y)sec(x+y)-xsecxy

Advertisements
Advertisements

Question

`lim_(y -> 0) ((x + y) sec(x + y) - x sec x)/y`

Sum
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)`

shaalaa.com
  Is there an error in this question or solution?
Chapter 13: Limits and Derivatives - Exercise [Page 241]

APPEARS IN

NCERT Exemplar Mathematics Exemplar [English] Class 11
Chapter 13 Limits and Derivatives
Exercise | Q 47 | Page 241

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

Evaluate the following limit.

`lim_(x → 0) x sec x`


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)[(1 - cos("n"x))/(1 - cos("m"x))]`


Evaluate the following limit :

`lim_(x -> pi) [(sqrt(1 - cosx) - sqrt(2))/(sin^2 x)]`


Select the correct answer from the given alternatives.

`lim_(x -> 0) ((5sinx - xcosx)/(2tanx - 3x^2))` =


Select the correct answer from the given alternatives.

`lim_(x -> pi/2) [(3cos x + cos 3x)/(2x - pi)^3]` =


Evaluate the following :

`lim_(x -> 0)[(secx^2 - 1)/x^4]`


Evaluate the following :

`lim_(x -> 0) [(x(6^x - 3^x))/(cos (6x) - cos (4x))]`


Evaluate the following :

`lim_(x -> pi/4) [(sinx - cosx)^2/(sqrt(2) - sinx - cosx)]`


Evaluate `lim_(x -> 0) (sqrt(2 + x) - sqrt(2))/x`


Evaluate `lim_(x -> pi/2) (secx - tanx)`


Evaluate `lim_(x -> pi/6) (2sin^2x + sin x - 1)/(2sin^2 x - 3sin x + 1)`


Evaluate `lim_(x -> a) (sqrt(a + 2x) - sqrt(3x))/(sqrt(3a + x) - 2sqrt(x))`


`lim_(x -> 0) sinx/(x(1 + cos x))` is equal to ______.


`lim_(x -> 0) |x|/x` is equal to ______.


If f(x) = x sinx, then f" `pi/2` is equal to ______.


Evaluate: `lim_(x -> sqrt(2)) (x^4 - 4)/(x^2 + 3sqrt(2x) - 8)`


Evaluate: `lim_(x -> 1) (x^7 - 2x^5 + 1)/(x^3 - 3x^2 + 2)`


Evaluate: `lim_(x -> 0) (sqrt(1 + x^3) - sqrt(1 - x^3))/x^2`


Evaluate: `lim_(x -> 1/2) (8x - 3)/(2x - 1) - (4x^2 + 1)/(4x^2 - 1)`


Evaluate: `lim_(x -> 0) (sin^2 2x)/(sin^2 4x)`


Evaluate: `lim_(x -> 0) (1 - cos 2x)/x^2`


Evaluate: `lim_(x -> pi/6) (sqrt(3) sin x - cos x)/(x - pi/6)`


Evaluate: `lim_(x -> a) (sin x - sin a)/(sqrt(x) - sqrt(a))`


Evaluate: `lim_(x -> 0) (sqrt(2) - sqrt(1 + cos x))/(sin^2x)`


`lim_(x -> pi/4) (tan^3x - tan x)/(cos(x + pi/4))`


`lim_(x -> pi) (1 - sin  x/2)/(cos  x/2 (cos  x/4 - sin  x/4))`


`lim_(x -> pi) sinx/(x - pi)` is equal to ______.


`lim_(x -> 1) (x^m - 1)/(x^n - 1)` is ______.


`lim_(x -> 0) sinx/(sqrt(x + 1) - sqrt(1 - x)` is ______.


If `f(x) = {{:(x^2 - 1",", 0 < x < 2),(2x + 3",", 2 ≤ x < 3):}`, the quadratic equation whose roots are `lim_(x -> 2^-) f(x)` and `lim_(x -> 2^+) f(x)` is ______. 


`lim_(x -> 3^+) x/([x])` = ______.


Let Sk = `sum_(r = 1)^k tan^-1(6^r/(2^(2r + 1) + 3^(2r + 1)))`. Then `lim_(k→∞)` Sk = is equal to ______.


If `lim_(n→∞)sum_(k = 2)^ncos^-1(1 + sqrt((k - 1)(k + 2)(k + 1)k)/(k(k + 1))) = π/λ`, then the value of λ is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×