I have to calculate the following limit lim x → + ∞ x ln ⁡ x 1 − sin ⁡ x

Question

I have to calculate the following limit      lim          x      →      +      ∞                  x      ln      ⁡      x              1      −      sin      ⁡      x

Dalton Lester · Accepted Answer

lim          n      →      ∞            n          2      +      ±      (      −      1              )        n            is infinity even though the denominator does not converge.The real problem with your function is that it isn’t defined for all of       R    . But if f is a function defined on a subset of       R   with no real upper bound, we can still define       lim          x      →      ∞        f  (  x  )  .In your case, you easily get, where f is defined, and   x  &amp;gt;  1  ,  f  (  x  )  =            x      ln      ⁡      x              1      −      sin      ⁡      x        ≥            x      ln      ⁡      x        2    ,since   0  &amp;lt;  1  −  sin  ⁡  x  ≤  2 in the domain of fSince             x      ln      ⁡      x        2    →  ∞  , this means   f  (  x  )  →  ∞  .

racodelitusmn · Accepted Answer

The quotient                     x        ln        ⁡        (        x        )                    1        −        sin        ⁡        (        x        )             is undefined when   x  ∈            π      2        +  2  π      Z  . Otherwise,   1  −  sin  ⁡  (  x  )  ⩽  2, and therefore                    (1)                                          x            ln            ⁡            (            x            )                                1            −            sin            ⁡            (            x            )                          ⩾                              x            ln            ⁡            (            x            )                    2                .            Since      lim          x      →      ∞        x  ln  ⁡  (  x  )  =  ∞  ,it follows from (1) that       lim          x      →      ∞                  x      ln      ⁡      (      x      )              1      −      sin      ⁡      (      x      )        =  ∞too.

I have to calculate the following limit <munder> <mo movablelimits="true" form="prefix">lim

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