I have to evaluate the following without L'Hopital's rule lim x → ∞ x tan ⁡ ( 1 / x )I can simplify this to be lim x → ∞ x sin ⁡ ( 1 / x )because lim x → ∞ cos ⁡ ( 1 / x ) = 1However, after that, I'm totally lost. L'Hopital's rule seems like my only option. Can someone help me out?

Question

I have to evaluate the following without L&#039;Hopital&#039;s rule      lim          x      →      ∞        x  tan  ⁡  (  1      /    x  )I can simplify this to be      lim          x      →      ∞        x  sin  ⁡  (  1      /    x  )because      lim          x      →      ∞        cos  ⁡  (  1      /    x  )  =  1However, after that, I&#039;m totally lost. L&#039;Hopital&#039;s rule seems like my only option. Can someone help me out?

Elias Flores · Accepted Answer

If you are allowed to use the fact that       lim          t      →      0                  sin      ⁡      (      t      )        t    =  1, then setting   t  =      1    x   yields      lim          x      →      ∞        x  sin  ⁡            (            1    x              )        =      lim          x      →      ∞                  sin      ⁡      (              1        x            )              1      x        =      lim          t      →      0                  sin      ⁡      (      t      )        t    =  1

I have to evaluate the following without L'Hopital's rule <munder> <mo movablelimits="tru

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