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Evaluate: limx→3x3+27x5+243

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प्रश्न

Evaluate: `lim_(x -> 3) (x^3 + 27)/(x^5 + 243)`

योग
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उत्तर

Given that `lim_(x -> 3) (x^3 + 27)/(x^5 + 243)`

= `lim_(x -> 3) ((x^3 + (3)^3)/(x - 3))/((x^5 + (3)^5)/(x - 3))`  ......[Dividing the Nr and Den. by x – 3]

= `(lim_(x -> 3) ((x^3 - (-3)^3)/(x + 3)))/(lim_(x -> 3) ((x^5 - (-3)^2)/(x + 3))`   ......`[lim_(x -> a) (f(x))/(g(x)) = (lim_(x -> a) f(x))/(lim_(x -> a) g(x))]`

= `(3(-3)^(3-1))/(5(-3)^(5 - 1))`

= `(3 xx (-3)^2)/(5 xx (-3)^4)`

= `1/(5 xx 3)`

= `1/15`

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अध्याय 13: Limits and Derivatives - Exercise [पृष्ठ २४०]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 11
अध्याय 13 Limits and Derivatives
Exercise | Q 12 | पृष्ठ २४०

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