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