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E v a l u a t e space limit as x rightwards arrow 0 of open parentheses 1 plus 2 x close parentheses to the power of 5 over x end exponent

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Asked by Topperlearning User 16th November 2016, 4:26 AM
Answered by Expert
Answer:

limit as x rightwards arrow 0 of open parentheses 1 plus 2 x close parentheses to the power of 5 over x end exponent
equals e to the power of limit as x rightwards arrow 0 of 5 over x log open parentheses 1 plus 2 x close parentheses end exponent
to the power of limit as x rightwards arrow 0 of 5 over x log open parentheses 1 plus 2 x close parentheses end exponent
equals 5 limit as x rightwards arrow 0 of fraction numerator 2 log left parenthesis 1 plus 2 x right parenthesis over denominator 2 x end fraction
equals 10
e to the power of limit as x rightwards arrow 0 of 5 over x log open parentheses 1 plus 2 x close parentheses end exponent equals e to the power of 10

Answered by Expert 16th November 2016, 6:26 AM
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