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प्रश्न
If f(x) = `(x^7 - 128)/(x^5 - 32)`, then find `lim_(x-> 2)` f(x)
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उत्तर
`lim_(x-> 2)` f(x)
= `lim_(x-> 2) (x^7 - 128)/(x^5 - 32)`
= `lim_(x-> 2) (x^7 - 2^7)/(x^5 - 2^5)`
= `(lim_(x-> 2) (x^7 - 2^7)/(x-2))/(lim_(x-> 2)(x^5 - 2^5)/(x-2))` ....[Divide both numerator amd denominator by x - 2]
`= (7 * 2^6)/(5 * 2^4)` ....`[lim_(x->"a") (x^n - "a"^n)/(x - a)]`
`= 7/5 xx 2^2 = 28/5`
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