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Question
Evaluate the following limit :
`lim_(x -> 0) [(cos("a"x) - cos("b"x))/(cos("c"x) - 1)]`
Solution
`lim_(x -> 0) (cos("a"x) - cos("b"x))/(cos("c"x) - 1)`
= `lim_(x -> 0) (-cos("a"x) + cos("b"x))/(1 - cos("c"x))`
= `lim_(x -> 0) (1 - cos "a"x - 1 + cos "b"x)/(1 - cos "c"x)`
= `lim_(x -> 0) ((1 - cos"a"x) - (1 - cos"b"x))/(1 - cos"c"x)`
= `lim_(x -> 0) (2sin^2 (("a"x)/2) - 2sin^2 ("b"x)/2)/(2sin^2 ("c"x)/2`
= `lim_(x -> 0) ((sin^2 (("a"x)/2) - sin^2 (("b"x)/2))/(x^2))/((sin^2 (("c"x)/2))/(x^2)) ...[("Divide numerator and denominator by" x^2),(because x -> 0"," therefore x ≠ 0 therefore x^2 ≠ 0)]`
= `(lim_(x -> 0) [(sin^2 (("a"x)/2))/x^2 - (sin^2(("b"x)/2))/x^2])/(lim_(x -> 0) (sin^2 (("c"x)/2))/x^2)`
= `(lim_(x -> 0) [(sin ("a"x)/2)/x]^2 - lim_(x -> 0) [(sin ("b"x)/2)/x]^2)/(lim_(x -> 0) [[sin ("c"x)/2)/x]^2`
= `(lim_(x -> 0) [(sin ("a"x)/2)/(("a"x)/2)]^2 * ("a"/2)^2 - lim_(x -> 0) [(sin ("b"x)/2)/(("b"x)/2)]^2 * ("b"/2)^2)/(lim_(x -> 0) [(sin ("c"x)/2)/(("c"x)/2)]^2 * ("c"/2)^2`
= `((1)^2 * "a"^2/4 - (1)^2 * "b"^2/4)/((1)^2 * "c"^2/4`
= `("a"^2/4 - "b"^2/4)/("c"^2/4)`
= `("a"^2 - "b"^2)/"c"^2`
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