Question

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Answer 1

what is this for

Answer 2

Try to do it if you feel like it

Answer 3

can you actually HELP ME ON SOME problems
please

Answer 4

Are they posted?

Answer 5

Note that the limit is of the form \[1^\infty \]
therefore we choose the approach to get it into the form of \[\large \lim_{n \to \infty} \left( 1 + \frac{1}{n}\right)^n =e \]
we first consider the expression within the parenthesis \[\Large 1 + \frac{5}{t}+ \frac{10}{t^4}=1 + \frac{5t^3+10}{t^4}= 1+ \frac{1}{\frac{t^4}{5t^3+10}} \] and use this to obtain:
\[\large \lim_{t \to \infty} \left(1+\frac{1}{\frac{t^4}{5t^3+10}} \right)^{\frac{t^4}{5t^3+10}^{\frac{5t^3+10}{t^4}^{9t}}}=\lim_{t \to \infty} e^{\frac{45t^4-90t}{t^4}} \]
we can take the limit inside because \(\exp\) is a continuous function and study the exponent instead which gives due to it's asymptotic behavior \(45\) \[\large \lim_{t \to \infty} \frac{45t^4-90t}{t^4} \sim \frac{45t^4}{t^4}=45\]
and obtain the result of \[\large \lim_{t \to \infty} \left(1 + \frac{5}{t}+\frac{10}{t^4} \right)^{9t}=e^{45} \]

Answer 6

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