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Evaluate the limit using an appropriate substitution. https://i.gyazo.com/4e72308519ac01411b2404ba5d2f260a.png Hint: [Hint: t = ln x]
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\[\lim_{lnx \rightarrow infty}\frac{ \ln2 + lnx }{ \ln3 + lnx } \] \[\lim_{lnx \rightarrow \infty}\frac{ \ln2/lnx + lnx/lnx }{ \ln3/lnx + lnx/lnx } \] = 1/1 = 1
how did you get that answer? may you explain the steps?
I did as the hint suggested , separate ln x from the coefficient . lnab = ln a + ln b meanwhile: ln2x = ln2 + lnx , ln3x = ln3 + lnx let lnx = t \[\lim_{lnx \rightarrow \ln \infty} = \lim_{t \rightarrow \infty} \frac{ \ln2 + t }{ \ln3 + t }\] ln2 , ln3 are just normal numbers. solve it like any other infinity limit.
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