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प्रश्न
Evaluate the following limits:
`lim_(alpha -> 0) (sin(alpha^"n"))/(sin alpha)^"m"`
उत्तर
We know `lim_(x -> 0) (sin x)/x` = 1
`lim_(alpha -> 0) (sin(alpha^"n"))/(sin alpha)^"m" = lim_(alpha -> 0) (sin (alpha^"n"))/(1/alpha^"n" (alpha^"n")) xx (alpha^(1/"m") * alpha^"m")/(sin alpha)^"m"`
= `lim_(alpha -> 0) alpha^"n" * (sin(alpha^"n"))/alpha^"n" xx 1/alpha^"m" * 1/(((sinalpha)^"m")/(alpha^"m"))`
= `lim_(alpha -> 0) alpha^("n" - "m") *(sin(alpha^"n"))/alpha^"n" xx 1/((sinalpha)/alpha)^"m"`
= `(lim_(alpha -> 0) alpha^("n" - "m")) xx (lim_(alpha^"n" -> 0) (sin(alpha^"n"))/alpha^"n") xx 1/(lim_(alpha -> 0) ((sin alpha)/alpha)^"m")`
Case (i) m = n
`lim_(alpha -> 0) (sin(alpha^"n"))/(sin alpha)^"m" = (lim_(alpha -> 0) alpha^("m" - "m")) (lim_(alpha^"m" -> 0) (sin(alpha^"m"))/alpha^"m") xx 1/(lim_(alpha -> 0) ((sin alpha)/alpha)^"m")`
= `(lim_(alpha -> 0) alpha^0) xx 1 xx 1/1`
`lim_(alpha -> 0) (sin(alpha^"n"))/(sin alpha)^"m" = 1 xx 1 xx 1` = 1
Case (ii) m > n then n – m < 0
`lim_(alpha -> 0) (sin(alpha^"n"))/(sin alpha)^"m" = (lim_(alpha -> 0) alpha^("n" - "m")) (lim_(apha^"m" -> 0) (sin(alpha^"m"))/alpha^"m") xx 1/((lim_(alpha-> 0) ((sin alpha)/alpha)^"m"))`
= `lim_(alpha -> 0) 1/(alpha^("m" - "n")) xx 1 xx 1/1`
= `oo xx 1 xx 1 = oo`
Since `lim_(alpha -> 0) 1/(alpha^("m" - "n")) = lim_(alpha -> 0) (1/0)^("m" - "n") = oo`
Case (iii) m < n then n – m > 0
`lim_(alpha -> 0) (sin(alpha^"n"))/(sin alpha)^"m" = (lim_(alpha -> 0) alpha^("n" - "m")) (lim_(alpha^"m" -> 0) (sin(alpha^"m"))/(alpha^"m")) xx 1/(lim_(alpha-> 0) ((sin alpha)/alpha)^"m"`
= `(0)^("n" - "m") xx 1 xx 1`
= 0
∴ `lim_(alpha -> 0) (sin(alpha^"n"))/(sin alpha)^"m" = {{:(1, "if", "m" = "n"),(oo, "if", "m" > "n"),(0, "if", m < n):}`
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