here's a limit that I'm struggling to calculate.

Averie Ferguson

Averie Ferguson

Answered question

2022-03-24

here's a limit that I'm struggling to calculate.
limx0x2lnxsinx(ex1)

Answer & Explanation

ostijum8dd

ostijum8dd

Beginner2022-03-25Added 7 answers

For x>0,
x2ln(x)sin(x)(ex1)=xsin(x)1xex11ln(x)

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