If lim x → 0 n x l n ( 1 + x + x 2 + ⋯ + x n ) exists and equals 5 1 , find n .

Using ln ( 1 + u ) ∼ u and x + x 2 + ⋯ + x n ∼ x , the limit is 1/ n . Hence 1/ n = 1/5 , so n = 5 .

Mathematics - Limits Question with Solution | TestHub

MathematicsLimitsValue of unknownMedium2 minQB

MathematicsMediumsingle choice

If $lim_{x \to 0} \frac{l n ( 1 + x + x ^{2} + \dots + x ^{n} )}{n x}$ exists and equals $\frac{1}{5}$ , find $n$ .

Answer:

Solution:

Using $ln (1 + u) \sim u$ and $x + x^{2} + \dots + x^{n} \sim x$ , the limit is $1/ n$ . Hence $1/ n = 1/5$ , so $n = 5$ .

Stream:JEESubject:MathematicsTopic:LimitsSubtopic:Value of unknown

⏱ 2mℹ️ Source: QB

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