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