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Evaluate the following limit : limx→0[1-cos(nx)1-cos(mx)]

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प्रश्न

Evaluate the following limit :

`lim_(x -> 0)[(1 - cos("n"x))/(1 - cos("m"x))]`

योग
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उत्तर

`lim_(x -> 0)(1 - cos"n"x)/(1 - cos"m"x)`

= `lim_(x -> 0) (1 - cos"n"x)/(1 - cos"m"x) xx (1 + cos "n"x)/(1 + cos "n"x) xx (1 + cos"m"x)/(1 + cos"m"x)`

= `lim_(x -> 0) ((1 - cos^2"n"x)(1 + cos"m"x))/((1 - cos^2"m"x)(1 + cos "n"x))`

= `lim_(x -> 0) (sin^2"n"x(1 + cos "m"x))/(sin^2"m"x(1 + cos "n"x))`

= `lim_(x -> 0) (((sin^2"n"x)/("n"^2x^2))(1 + cos "m"x))/(((sin^2"m"x)/("m"^2x^2))(1 + cos "n"x)) xx "n"^2/"m"^2`  ...[∵ x → 0, x ≠ 0 ∴ x2 ≠ 0]

= `"n"^2/"m"^2 ([lim_(x -> 0) (sin"n"x)/("n"x)]^2 xx [lim_(x -> 0) (1 + cos "m"x)])/([lim_(x -> 0) (sin"m"x)/("m"x)]^2 xx [lim_(x -> 0) (1 + cos "n"x)])`

= `"n"^2/"m"^2 (1^2*(1 + cos 0))/(1^2*(1 + cos 0))  ...[because  x -> 0  therefore "m"x, "n"x -> 0 and lim_(theta -> 0)  sintheta/theta = 1]`

= `"n"^2/"m"^2`

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अध्याय 7: Limits - Exercise 7.4 [पृष्ठ १४८]

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