मराठी

Evaluate: limx→1(1+x)6-1(1+x)2-1 - Mathematics

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प्रश्न

Evaluate: `lim_(x -> 1) ((1 + x)^6 - 1)/((1 + x)^2 - 1)`

बेरीज
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उत्तर

Given that `lim_(x -> 0) ((1 + x)^6 - 1)/((1 + x)^2 - 1)`

Dividing the numerator and denominator by x, we get

= `lim_(x -> 0) (((1 + x)^6 - 1)/x)/(((1 + x)^2 - 1)/x)`

Putting 1 + x = y

⇒ x = y – 1

= `lim_((y - 1 -> 0),(because y -> 1))  ((y^6 - (1)^6)/(y - 1))/((y^2 - (1)^2)/(y - 1))`

= `(lim_(y -> 1) (y^6 - (1)^6)/(y - 1))/(lim_(y -> 1) (y^2 - (1)^2)/(y - 1))`   .....`[lim_(x -> a) (f(x))/(g(x)) = (lim_(x -> a) f(x))/(lim_(x -> a) g(x))]`

= `(6 * (1)^(6 - 1))/(2 * (1)^(2 - 1))`

= `6/2`

= 3   ......`["Using" lim_(x -> a) (x^n - a^n)/(x - a) = n * a^(n - 1)]`

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पाठ 13: Limits and Derivatives - Exercise [पृष्ठ २३९]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
पाठ 13 Limits and Derivatives
Exercise | Q 5 | पृष्ठ २३९

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