Evaluating limn→∞(1+sin⁡n5n+1)2n+3 (a 1∞ indeterminate form)

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karton · Accepted Answer

Take logarithm and write it as 25((5n+1+132)ln⁡(1+sin⁡n5n+1)) the limit becomes =25limn→+∞ln⁡(1+sin⁡n5n+1)sin⁡n5n+1sin⁡n =limn→+∞sin⁡n which does not exist. Observe that limn→+∞135ln⁡(1+sin⁡n5n+1)=0 So, your limit does not exist.

user_27qwe · Accepted Answer

Define L:=(1+sin⁡n5n+1)2n+3 and take the logarithm of both sides to bring the exponent down. Proceed from there.

Vasquez · Accepted Answer

Using 1+rx&amp;lt;(1+x)r&amp;lt;exp(rx), we have 1+2n+35n+1sin⁡n&amp;lt;(1+sin⁡n5n+1)2n+3&amp;lt;exp(2n+35n+1sin⁡n) For any n such that sin⁡n&amp;gt;12 (and there are infinitely many), we have 1+2n+35n+1sin⁡n&amp;gt;1+25⋅12=1.2 For any n such that sin⁡n&amp;lt;−12 (also infinitely many), we have exp(2n+35n+1sin⁡n)&amp;lt;exp(−25⋅12)&amp;lt;0.9 And thus the limit can&#039;t exist, as the value is infinitely often less than 0.9 and infinitely often greater than 1.2.

Evaluating \lim_{n \to \infty} (1+\frac{\sin n}{5n+1})^{2n+3} (a 1^{\infty} indeterminate form)

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