Limit of a natural log function












0














Lim x-->0 ln((x^2)/16-x^2)



How can I take the limit of this? I can't apply L'Hopital because only the numerator is indeterminate. Thanks!






limits







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  • 4




    Where's the problem? This is not indeterminate. The argument of the log tends to $0^+$, hence the log tends to $-infty$.
    – Bernard
    Nov 20 '18 at 0:05










  • @bernard try placing a backslash before infty.
    – Oscar Lanzi
    Nov 20 '18 at 1:54
















0














Lim x-->0 ln((x^2)/16-x^2)



How can I take the limit of this? I can't apply L'Hopital because only the numerator is indeterminate. Thanks!






limits







share|cite|improve this question


















  • 4




    Where's the problem? This is not indeterminate. The argument of the log tends to $0^+$, hence the log tends to $-infty$.
    – Bernard
    Nov 20 '18 at 0:05










  • @bernard try placing a backslash before infty.
    – Oscar Lanzi
    Nov 20 '18 at 1:54














0












0








0







Lim x-->0 ln((x^2)/16-x^2)



How can I take the limit of this? I can't apply L'Hopital because only the numerator is indeterminate. Thanks!






limits







share|cite|improve this question













Lim x-->0 ln((x^2)/16-x^2)



How can I take the limit of this? I can't apply L'Hopital because only the numerator is indeterminate. Thanks!






limits




limits






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share|cite|improve this question










asked Nov 20 '18 at 0:01









M Do

124




124








  • 4




    Where's the problem? This is not indeterminate. The argument of the log tends to $0^+$, hence the log tends to $-infty$.
    – Bernard
    Nov 20 '18 at 0:05










  • @bernard try placing a backslash before infty.
    – Oscar Lanzi
    Nov 20 '18 at 1:54














  • 4




    Where's the problem? This is not indeterminate. The argument of the log tends to $0^+$, hence the log tends to $-infty$.
    – Bernard
    Nov 20 '18 at 0:05










  • @bernard try placing a backslash before infty.
    – Oscar Lanzi
    Nov 20 '18 at 1:54








4




4




Where's the problem? This is not indeterminate. The argument of the log tends to $0^+$, hence the log tends to $-infty$.
– Bernard
Nov 20 '18 at 0:05




Where's the problem? This is not indeterminate. The argument of the log tends to $0^+$, hence the log tends to $-infty$.
– Bernard
Nov 20 '18 at 0:05












@bernard try placing a backslash before infty.
– Oscar Lanzi
Nov 20 '18 at 1:54




@bernard try placing a backslash before infty.
– Oscar Lanzi
Nov 20 '18 at 1:54










1 Answer
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The answer is $-infty$. We have



$$lim_{xto 0} frac{ln(x^{2})}{16 - x^{2}}.$$
When $xto 0$, $ln(x^2) rightarrow -infty$. Also, $16-x^{2} rightarrow 16$.



Therefore, the answer is $-infty$.






share|cite|improve this answer





















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    1 Answer
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    1 Answer
    1






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    The answer is $-infty$. We have



    $$lim_{xto 0} frac{ln(x^{2})}{16 - x^{2}}.$$
    When $xto 0$, $ln(x^2) rightarrow -infty$. Also, $16-x^{2} rightarrow 16$.



    Therefore, the answer is $-infty$.






    share|cite|improve this answer


























      2














      The answer is $-infty$. We have



      $$lim_{xto 0} frac{ln(x^{2})}{16 - x^{2}}.$$
      When $xto 0$, $ln(x^2) rightarrow -infty$. Also, $16-x^{2} rightarrow 16$.



      Therefore, the answer is $-infty$.






      share|cite|improve this answer
























        2












        2








        2






        The answer is $-infty$. We have



        $$lim_{xto 0} frac{ln(x^{2})}{16 - x^{2}}.$$
        When $xto 0$, $ln(x^2) rightarrow -infty$. Also, $16-x^{2} rightarrow 16$.



        Therefore, the answer is $-infty$.






        share|cite|improve this answer












        The answer is $-infty$. We have



        $$lim_{xto 0} frac{ln(x^{2})}{16 - x^{2}}.$$
        When $xto 0$, $ln(x^2) rightarrow -infty$. Also, $16-x^{2} rightarrow 16$.



        Therefore, the answer is $-infty$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 20 '18 at 1:26









        Ekesh

        5326




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