A proof of why y=mx grows faster than y = ln ⁡ x for all m ≠ 0How do I prove that the linear function y=mx grows faster than the logarithmic function?

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A proof of why y=mx grows faster than   y  =  ln  ⁡  x for all   m  ≠  0How do I prove that the linear function y=mx grows faster than the logarithmic function?

t5an1izqp · Accepted Answer

Alternatively, you could prove as follows:      lim          x      →      ∞                  m      x              ln      ⁡      (      x      )        =      lim          x      →      ∞                  m      x              2      ln      ⁡      (              x            )        ≥      lim          x      →      ∞                  m      x              2              x              =      lim          x      →      ∞                  m              x              2    =  ∞  ,  (  m  ≠  0  )which takes advantage of   ln  ⁡  (  x  )  &amp;lt;      x

Juan Leonard · Accepted Answer

To prove this, consider the limit:  L  =      lim          x      →      ∞                  m      x              ln      ⁡      x      (m&amp;gt;0). Using, for instance, L&#039;Hospital&#039;s rule, you can easily see that:  L  =  ∞This means that   f  (  x  )  =  m  x grows faster than   g  (  x  )  =  ln  ⁡  x

A proof of why y=mx grows faster than y=ln x for all m !=0 How do I prove that the linear function y=mx grows faster than the logarithmic function?

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