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प्रश्न
Test the continuity of the following function at the point or interval indicated against them :
f(x) `{:(= ((27 - 2x)^(1/3) - 3)/(9 - 3(243 + 5x)^(1/5))",", "for" x ≠ 0),(= 2",", "for" x = 0):}}` at x = 0.
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उत्तर
f(0) = 2 ...(Given)
`lim_(x -> 0) "f"(x) = lim_(x -> 0) ((27 - 2x)^(1/3) - 3)/(9 - 3(243 + 5x)^(1/5))`
= `lim_(x -> 0) ((27 - 2x)^(1/3) - 3)/(-3[(243 + 5x)^(1/5) - 3]`
= `(-1)/(3) lim_(x -> 0) ((27 - 2x)^(1/3) - (27)^(1/3))/((243 + 5x)^(1/5) - (243)^(1/5))`
= `(-1)/(3) lim_(x -> 0) (((27 - 2x)^(1/3) - 27^(1/3))/((27 - 2x) - 27) xx [(27 - 2x) - 27])/(((243 + 5x)^(1/5) - (243)^(1/5))/((243 + 5x) - 243) xx [(243 + 5x) - 243])`
...`[("As" x -> 0"," -2x -> 0 and 5x -> 0),(therefore (27 - 2x) - 27 -> 0 and (243 + 5x) - 243 -> 0),(therefore (27 - 2x) - 27 ≠ 0 and (243 + 5x) - 243 ≠ 0)]`
= `(-1)/(3) (lim_(x -> 0) ((27 - 2x)^(1/3) - 27^(1/3))/((27 - 2x) - 27) xx (-2x))/(lim_(x -> 0) ((243 + 5x)^(1/3) - (243)^(1/5))/((243 + 5x) - 243) xx (5x)`
= `(-1)/(3) xx (-2)/(5) xx (lim_(x -> 0)((27 - 2x)^(1/3) - 27^(1/3))/((27 - 2x) - 27))/(lim_(x -> 0) ((243 + 5x)^(1/5) - (243)^(1/5))/((243 + 5x) - 243)` ...[∵ x → 0, x ≠ 0]
= `2/15 xx (1/3(27)^((-2)/3))/(1/5(243)^((-4)/5)) ...[because lim_(x -> "a") (x^"n" - "a"^"n")/(x - "a") = "na"^("n" - 1)]`
= `2/15 xx 5/3 xx ((3^3)^((-2)/3))/((3^5)^((-4)/5))`
= `2/9 xx (3)^(-2)/(3)^(-4)`
= `2/9 xx (3)^2`
= 2
∴ `lim_(x -> 0) "f"(x)` = f(0)
∴ f(x) is continuous at x = 0
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