Could somebody validate my proof regarding the limit of ln ⁡ ( x n ) when, x n tends to a?So, let me cearly state the problem:Let ( x n ) be a convergent sequence, with: x n &gt; 0, ∀ n, n natural number, and x n → a, with a &gt; 0. Then ln ⁡ x n → ln ⁡ a Here is my idea for a proof:Our goal is to proof that there ∀ ϵ &gt; 0 there is some n ϵ , such that ∀ n ≥ n ϵ , we have that | ln ⁡ x n − ln ⁡ a | &lt; ϵSo here is what I did. First: | ln ⁡ x n − ln ⁡ a | = | ln ⁡ x n a | Then, beacause ln ⁡ x &lt; x, ∀ x &gt; 0, it easily follows that | ln ⁡ x | &lt; x, ∀ x &gt; 0. Applying this, we have that: | ln ⁡ x n − ln ⁡ a | = | ln ⁡ x n a | &lt; x n a = x n − a + a a = x n − a a + 1Now, we use the basic property of the absolute value: x n − a ≤ | x n − a | , that gives us: | ln ⁡ x n − ln ⁡ a | &lt; x n − a a + 1 ≤ | x n − a | a + 1Now, we use the fact that x n → a. So, ∀ ϵ &gt; 0 there is some n ϵ , such that ∀ n ≥ n ϵ , we have that | x n − a | &lt; ϵ. We choose an epsilon that takes the form a ( ϵ 0 − 1 ). This choice is possible for any ϵ 0 Now, we have managed to obtain that: | ln ⁡ x n − ln ⁡ a | &lt; a ( ϵ 0 − 1 ) a + 1 = ϵ 0 , ∀ n ≥ n ϵ Since ∀ ϵ 0 &gt; 0 we can find an n ϵ , such that ∀ n ≥ n ϵ , the above inequality is staisfied, our claim is proved.So, can you please tell me if my proof si correct? I've tried to find a proof, only for the limit of sequences! This problem has been on my nerves for s while. Also, probably there is some simpler way to do it, but I couldn't find it.

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Could somebody validate my proof regarding the limit of   ln  ⁡  (      x    n    ) when,       x    n   tends to   a?So, let me cearly state the problem:Let   (      x    n    ) be a convergent sequence, with:       x    n    &amp;gt;  0,   ∀  n, n natural number, and       x    n    →  a, with   a  &amp;gt;  0. Then   ln  ⁡            x      n        →  ln  ⁡      a  Here is my idea for a proof:Our goal is to proof that there   ∀  ϵ  &amp;gt;  0 there is some       n          ϵ      , such that   ∀  n  ≥      n          ϵ      , we have that       |    ln  ⁡            x      n        −  ln  ⁡      a        |    &amp;lt;  ϵSo here is what I did. First:      |    ln  ⁡            x      n        −  ln  ⁡      a        |    =      |    ln  ⁡                    x        n            a            |  Then, beacause   ln  ⁡      x    &amp;lt;  x,   ∀  x  &amp;gt;  0, it easily follows that          |    ln  ⁡      x        |    &amp;lt;  x,   ∀  x  &amp;gt;  0. Applying this, we have that:      |    ln  ⁡            x      n        −  ln  ⁡      a        |    =      |    ln  ⁡                    x        n            a            |    &amp;lt;            x      n        a    =                    x        n            −      a      +      a        a    =                    x        n            −      a        a    +  1Now, we use the basic property of the absolute value:       x    n    −  a  ≤      |        x    n    −  a      |  , that gives us:      |    ln  ⁡            x      n        −  ln  ⁡      a        |    &amp;lt;                    x        n            −      a        a    +  1  ≤                    |                    x        n            −      a              |              a    +  1Now, we use the fact that       x    n    →  a. So,   ∀  ϵ  &amp;gt;  0 there is some       n          ϵ      , such that   ∀  n  ≥      n          ϵ      , we have that       |        x    n    −  a      |    &amp;lt;  ϵ. We choose an epsilon that takes the form   a  (      ϵ    0    −  1  ). This choice is possible for any       ϵ    0  Now, we have managed to obtain that:      |    ln  ⁡            x      n        −  ln  ⁡      a        |    &amp;lt;            a      (              ϵ        0            −      1      )        a    +  1  =      ϵ    0    ,  ∀  n  ≥      n          ϵ      Since   ∀      ϵ    0    &amp;gt;  0 we can find an       n          ϵ      , such that   ∀  n  ≥      n          ϵ      , the above inequality is staisfied, our claim is proved.So, can you please tell me if my proof si correct? I&#039;ve tried to find a proof, only for the limit of sequences! This problem has been on my nerves for s while. Also, probably there is some simpler way to do it, but I couldn&#039;t find it.

Frances Dodson · Accepted Answer

Yes, it is correct.In fact the property      lim          n      →      ∞        f  (      x    n    )  =  f  (      lim          n      →      ∞            x    n    )you proved for   ln  ⁡  (  x  ) is called sequential continuty. It is equivalent to continuity, meaning any function   f is continous if and only if this holds for all series   (  x      )    n  So along the way you have shown that   ln  ⁡  (  x  ) is continuous on   (  0  ,  ∞  )

So, let me cearly state the problem: Let (x_n) be a convergent sequence, with: x_n>0, AA n, n natural number, and x_n->a, with a>0. Then lnx_n->lna.

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