Find $lim_{x to 0} (sin x)^{1/x} + (1/x)^{sin x}$












1












$begingroup$


Evaluate the limit $lim_{x to 0} (sin x)^{1/x} + (1/x)^{sin x}$



This limit is of the form $(0)^{infty} + {infty}^{0}$



I tried to calculate both limits individually, and got confused if the first term is an indeterminate form of not. If it is not, then this value is $0$ and second term is $1$.



But if first term is an indeterminate form then how would i calculate the limit?






calculus







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$endgroup$








  • 1




    $begingroup$
    Is $x$ meant to tend to $0$ rather than $infty$? Maybe $0^+$?
    $endgroup$
    – Olivier Moschetta
    Jan 3 at 14:33












  • $begingroup$
    I believe it should be $xto 0^+$ as pointed out by @OlivierMoschetta, otherwise the limit does not exist
    $endgroup$
    – roman
    Jan 3 at 14:54










  • $begingroup$
    Yea it's 0. Sorry for inconvenience.
    $endgroup$
    – Mathsaddict
    Jan 3 at 15:04
















1












$begingroup$


Evaluate the limit $lim_{x to 0} (sin x)^{1/x} + (1/x)^{sin x}$



This limit is of the form $(0)^{infty} + {infty}^{0}$



I tried to calculate both limits individually, and got confused if the first term is an indeterminate form of not. If it is not, then this value is $0$ and second term is $1$.



But if first term is an indeterminate form then how would i calculate the limit?






calculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Is $x$ meant to tend to $0$ rather than $infty$? Maybe $0^+$?
    $endgroup$
    – Olivier Moschetta
    Jan 3 at 14:33












  • $begingroup$
    I believe it should be $xto 0^+$ as pointed out by @OlivierMoschetta, otherwise the limit does not exist
    $endgroup$
    – roman
    Jan 3 at 14:54










  • $begingroup$
    Yea it's 0. Sorry for inconvenience.
    $endgroup$
    – Mathsaddict
    Jan 3 at 15:04














1












1








1


2



$begingroup$


Evaluate the limit $lim_{x to 0} (sin x)^{1/x} + (1/x)^{sin x}$



This limit is of the form $(0)^{infty} + {infty}^{0}$



I tried to calculate both limits individually, and got confused if the first term is an indeterminate form of not. If it is not, then this value is $0$ and second term is $1$.



But if first term is an indeterminate form then how would i calculate the limit?






calculus







share|cite|improve this question











$endgroup$




Evaluate the limit $lim_{x to 0} (sin x)^{1/x} + (1/x)^{sin x}$



This limit is of the form $(0)^{infty} + {infty}^{0}$



I tried to calculate both limits individually, and got confused if the first term is an indeterminate form of not. If it is not, then this value is $0$ and second term is $1$.



But if first term is an indeterminate form then how would i calculate the limit?






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 3 at 14:54







Mathsaddict

















asked Jan 3 at 14:06









MathsaddictMathsaddict

3669




3669








  • 1




    $begingroup$
    Is $x$ meant to tend to $0$ rather than $infty$? Maybe $0^+$?
    $endgroup$
    – Olivier Moschetta
    Jan 3 at 14:33












  • $begingroup$
    I believe it should be $xto 0^+$ as pointed out by @OlivierMoschetta, otherwise the limit does not exist
    $endgroup$
    – roman
    Jan 3 at 14:54










  • $begingroup$
    Yea it's 0. Sorry for inconvenience.
    $endgroup$
    – Mathsaddict
    Jan 3 at 15:04














  • 1




    $begingroup$
    Is $x$ meant to tend to $0$ rather than $infty$? Maybe $0^+$?
    $endgroup$
    – Olivier Moschetta
    Jan 3 at 14:33












  • $begingroup$
    I believe it should be $xto 0^+$ as pointed out by @OlivierMoschetta, otherwise the limit does not exist
    $endgroup$
    – roman
    Jan 3 at 14:54










  • $begingroup$
    Yea it's 0. Sorry for inconvenience.
    $endgroup$
    – Mathsaddict
    Jan 3 at 15:04








1




1




$begingroup$
Is $x$ meant to tend to $0$ rather than $infty$? Maybe $0^+$?
$endgroup$
– Olivier Moschetta
Jan 3 at 14:33






$begingroup$
Is $x$ meant to tend to $0$ rather than $infty$? Maybe $0^+$?
$endgroup$
– Olivier Moschetta
Jan 3 at 14:33














$begingroup$
I believe it should be $xto 0^+$ as pointed out by @OlivierMoschetta, otherwise the limit does not exist
$endgroup$
– roman
Jan 3 at 14:54




$begingroup$
I believe it should be $xto 0^+$ as pointed out by @OlivierMoschetta, otherwise the limit does not exist
$endgroup$
– roman
Jan 3 at 14:54












$begingroup$
Yea it's 0. Sorry for inconvenience.
$endgroup$
– Mathsaddict
Jan 3 at 15:04




$begingroup$
Yea it's 0. Sorry for inconvenience.
$endgroup$
– Mathsaddict
Jan 3 at 15:04










2 Answers
2






active

oldest

votes


















2












$begingroup$

I'm assuming $xrightarrow 0^+$ rather than $xrightarrowinfty$. The first limit is not an indeterminate form, but here is a direct calculation anyway.



Assume $0<x<1$ so that $1/x>1$. For every $ain ]0,1[$ we then have
$$0<a^{1/x}<a$$
(recall for instance that $a^3<a^2$ for such $a$), in particular if $a=sin(x)$:
$$0<sin(x)^{1/x}leqsin(x)$$
By the squeeze theorem we obtain
$$lim_{xrightarrow 0^+}sin(x)^{1/x}=0$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    First term is not indeterminate. Why? When $x to 0$ $sin x to 0$ and $1/x to infty$ this should be $0^{infty}$ indeterminate form.
    $endgroup$
    – Mathsaddict
    Jan 3 at 15:18










  • $begingroup$
    You can adapt my calculations to prove that if $f(x)rightarrow 0$ and $g(x)rightarrowinfty$ then $f(x)^{g(x)}rightarrow 0$. Essentially $g(x)>1$ and $|f(x)|<1$ if $x$ is small enough so that $|f(x)|^{g(x)}leq |f(x)|rightarrow 0$.
    $endgroup$
    – Olivier Moschetta
    Jan 3 at 15:25



















3












$begingroup$

$$
L = lim_{xto 0^+} left((sin x)^{1over x} + left({1over x}right)^{sin x}right) \
= lim_{xto 0^+} frac{(sin x)^{1over x}cdot x^{sin x} + 1}{x^{sin x}}
$$



By $sin x sim x$ as $xto 0$:
$$
frac{(sin x)^{1over x}cdot x^{sin x} + 1}{x^{sin x}} sim frac{x^{1over x}x^x + 1}{x^x} = frac{x^{x+{1over x}}+ 1}{x^x}
$$



By the fact that $lim_{xto 0^+} x^x = 1$:
$$
begin{align}
lim_{xto 0^+} frac{x^{x+{1over x}}+ 1}{x^x} &= lim_{xto 0^+} left(x^{x+{1over x}} + 1right) \
&= 1+ lim_{xto 0^+} x^x cdot x^{1over x} \
&= 1 + lim_{xto 0^+}x^x cdot lim_{xto 0^+} sqrt[x]{x}\
&= 1 + 1cdot 0 = 1
end{align}
$$






share|cite|improve this answer









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    2 Answers
    2






    active

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    2 Answers
    2






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    I'm assuming $xrightarrow 0^+$ rather than $xrightarrowinfty$. The first limit is not an indeterminate form, but here is a direct calculation anyway.



    Assume $0<x<1$ so that $1/x>1$. For every $ain ]0,1[$ we then have
    $$0<a^{1/x}<a$$
    (recall for instance that $a^3<a^2$ for such $a$), in particular if $a=sin(x)$:
    $$0<sin(x)^{1/x}leqsin(x)$$
    By the squeeze theorem we obtain
    $$lim_{xrightarrow 0^+}sin(x)^{1/x}=0$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      First term is not indeterminate. Why? When $x to 0$ $sin x to 0$ and $1/x to infty$ this should be $0^{infty}$ indeterminate form.
      $endgroup$
      – Mathsaddict
      Jan 3 at 15:18










    • $begingroup$
      You can adapt my calculations to prove that if $f(x)rightarrow 0$ and $g(x)rightarrowinfty$ then $f(x)^{g(x)}rightarrow 0$. Essentially $g(x)>1$ and $|f(x)|<1$ if $x$ is small enough so that $|f(x)|^{g(x)}leq |f(x)|rightarrow 0$.
      $endgroup$
      – Olivier Moschetta
      Jan 3 at 15:25
















    2












    $begingroup$

    I'm assuming $xrightarrow 0^+$ rather than $xrightarrowinfty$. The first limit is not an indeterminate form, but here is a direct calculation anyway.



    Assume $0<x<1$ so that $1/x>1$. For every $ain ]0,1[$ we then have
    $$0<a^{1/x}<a$$
    (recall for instance that $a^3<a^2$ for such $a$), in particular if $a=sin(x)$:
    $$0<sin(x)^{1/x}leqsin(x)$$
    By the squeeze theorem we obtain
    $$lim_{xrightarrow 0^+}sin(x)^{1/x}=0$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      First term is not indeterminate. Why? When $x to 0$ $sin x to 0$ and $1/x to infty$ this should be $0^{infty}$ indeterminate form.
      $endgroup$
      – Mathsaddict
      Jan 3 at 15:18










    • $begingroup$
      You can adapt my calculations to prove that if $f(x)rightarrow 0$ and $g(x)rightarrowinfty$ then $f(x)^{g(x)}rightarrow 0$. Essentially $g(x)>1$ and $|f(x)|<1$ if $x$ is small enough so that $|f(x)|^{g(x)}leq |f(x)|rightarrow 0$.
      $endgroup$
      – Olivier Moschetta
      Jan 3 at 15:25














    2












    2








    2





    $begingroup$

    I'm assuming $xrightarrow 0^+$ rather than $xrightarrowinfty$. The first limit is not an indeterminate form, but here is a direct calculation anyway.



    Assume $0<x<1$ so that $1/x>1$. For every $ain ]0,1[$ we then have
    $$0<a^{1/x}<a$$
    (recall for instance that $a^3<a^2$ for such $a$), in particular if $a=sin(x)$:
    $$0<sin(x)^{1/x}leqsin(x)$$
    By the squeeze theorem we obtain
    $$lim_{xrightarrow 0^+}sin(x)^{1/x}=0$$






    share|cite|improve this answer









    $endgroup$



    I'm assuming $xrightarrow 0^+$ rather than $xrightarrowinfty$. The first limit is not an indeterminate form, but here is a direct calculation anyway.



    Assume $0<x<1$ so that $1/x>1$. For every $ain ]0,1[$ we then have
    $$0<a^{1/x}<a$$
    (recall for instance that $a^3<a^2$ for such $a$), in particular if $a=sin(x)$:
    $$0<sin(x)^{1/x}leqsin(x)$$
    By the squeeze theorem we obtain
    $$lim_{xrightarrow 0^+}sin(x)^{1/x}=0$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 3 at 14:43









    Olivier MoschettaOlivier Moschetta

    2,8311411




    2,8311411












    • $begingroup$
      First term is not indeterminate. Why? When $x to 0$ $sin x to 0$ and $1/x to infty$ this should be $0^{infty}$ indeterminate form.
      $endgroup$
      – Mathsaddict
      Jan 3 at 15:18










    • $begingroup$
      You can adapt my calculations to prove that if $f(x)rightarrow 0$ and $g(x)rightarrowinfty$ then $f(x)^{g(x)}rightarrow 0$. Essentially $g(x)>1$ and $|f(x)|<1$ if $x$ is small enough so that $|f(x)|^{g(x)}leq |f(x)|rightarrow 0$.
      $endgroup$
      – Olivier Moschetta
      Jan 3 at 15:25


















    • $begingroup$
      First term is not indeterminate. Why? When $x to 0$ $sin x to 0$ and $1/x to infty$ this should be $0^{infty}$ indeterminate form.
      $endgroup$
      – Mathsaddict
      Jan 3 at 15:18










    • $begingroup$
      You can adapt my calculations to prove that if $f(x)rightarrow 0$ and $g(x)rightarrowinfty$ then $f(x)^{g(x)}rightarrow 0$. Essentially $g(x)>1$ and $|f(x)|<1$ if $x$ is small enough so that $|f(x)|^{g(x)}leq |f(x)|rightarrow 0$.
      $endgroup$
      – Olivier Moschetta
      Jan 3 at 15:25
















    $begingroup$
    First term is not indeterminate. Why? When $x to 0$ $sin x to 0$ and $1/x to infty$ this should be $0^{infty}$ indeterminate form.
    $endgroup$
    – Mathsaddict
    Jan 3 at 15:18




    $begingroup$
    First term is not indeterminate. Why? When $x to 0$ $sin x to 0$ and $1/x to infty$ this should be $0^{infty}$ indeterminate form.
    $endgroup$
    – Mathsaddict
    Jan 3 at 15:18












    $begingroup$
    You can adapt my calculations to prove that if $f(x)rightarrow 0$ and $g(x)rightarrowinfty$ then $f(x)^{g(x)}rightarrow 0$. Essentially $g(x)>1$ and $|f(x)|<1$ if $x$ is small enough so that $|f(x)|^{g(x)}leq |f(x)|rightarrow 0$.
    $endgroup$
    – Olivier Moschetta
    Jan 3 at 15:25




    $begingroup$
    You can adapt my calculations to prove that if $f(x)rightarrow 0$ and $g(x)rightarrowinfty$ then $f(x)^{g(x)}rightarrow 0$. Essentially $g(x)>1$ and $|f(x)|<1$ if $x$ is small enough so that $|f(x)|^{g(x)}leq |f(x)|rightarrow 0$.
    $endgroup$
    – Olivier Moschetta
    Jan 3 at 15:25











    3












    $begingroup$

    $$
    L = lim_{xto 0^+} left((sin x)^{1over x} + left({1over x}right)^{sin x}right) \
    = lim_{xto 0^+} frac{(sin x)^{1over x}cdot x^{sin x} + 1}{x^{sin x}}
    $$



    By $sin x sim x$ as $xto 0$:
    $$
    frac{(sin x)^{1over x}cdot x^{sin x} + 1}{x^{sin x}} sim frac{x^{1over x}x^x + 1}{x^x} = frac{x^{x+{1over x}}+ 1}{x^x}
    $$



    By the fact that $lim_{xto 0^+} x^x = 1$:
    $$
    begin{align}
    lim_{xto 0^+} frac{x^{x+{1over x}}+ 1}{x^x} &= lim_{xto 0^+} left(x^{x+{1over x}} + 1right) \
    &= 1+ lim_{xto 0^+} x^x cdot x^{1over x} \
    &= 1 + lim_{xto 0^+}x^x cdot lim_{xto 0^+} sqrt[x]{x}\
    &= 1 + 1cdot 0 = 1
    end{align}
    $$






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      $$
      L = lim_{xto 0^+} left((sin x)^{1over x} + left({1over x}right)^{sin x}right) \
      = lim_{xto 0^+} frac{(sin x)^{1over x}cdot x^{sin x} + 1}{x^{sin x}}
      $$



      By $sin x sim x$ as $xto 0$:
      $$
      frac{(sin x)^{1over x}cdot x^{sin x} + 1}{x^{sin x}} sim frac{x^{1over x}x^x + 1}{x^x} = frac{x^{x+{1over x}}+ 1}{x^x}
      $$



      By the fact that $lim_{xto 0^+} x^x = 1$:
      $$
      begin{align}
      lim_{xto 0^+} frac{x^{x+{1over x}}+ 1}{x^x} &= lim_{xto 0^+} left(x^{x+{1over x}} + 1right) \
      &= 1+ lim_{xto 0^+} x^x cdot x^{1over x} \
      &= 1 + lim_{xto 0^+}x^x cdot lim_{xto 0^+} sqrt[x]{x}\
      &= 1 + 1cdot 0 = 1
      end{align}
      $$






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        $$
        L = lim_{xto 0^+} left((sin x)^{1over x} + left({1over x}right)^{sin x}right) \
        = lim_{xto 0^+} frac{(sin x)^{1over x}cdot x^{sin x} + 1}{x^{sin x}}
        $$



        By $sin x sim x$ as $xto 0$:
        $$
        frac{(sin x)^{1over x}cdot x^{sin x} + 1}{x^{sin x}} sim frac{x^{1over x}x^x + 1}{x^x} = frac{x^{x+{1over x}}+ 1}{x^x}
        $$



        By the fact that $lim_{xto 0^+} x^x = 1$:
        $$
        begin{align}
        lim_{xto 0^+} frac{x^{x+{1over x}}+ 1}{x^x} &= lim_{xto 0^+} left(x^{x+{1over x}} + 1right) \
        &= 1+ lim_{xto 0^+} x^x cdot x^{1over x} \
        &= 1 + lim_{xto 0^+}x^x cdot lim_{xto 0^+} sqrt[x]{x}\
        &= 1 + 1cdot 0 = 1
        end{align}
        $$






        share|cite|improve this answer









        $endgroup$



        $$
        L = lim_{xto 0^+} left((sin x)^{1over x} + left({1over x}right)^{sin x}right) \
        = lim_{xto 0^+} frac{(sin x)^{1over x}cdot x^{sin x} + 1}{x^{sin x}}
        $$



        By $sin x sim x$ as $xto 0$:
        $$
        frac{(sin x)^{1over x}cdot x^{sin x} + 1}{x^{sin x}} sim frac{x^{1over x}x^x + 1}{x^x} = frac{x^{x+{1over x}}+ 1}{x^x}
        $$



        By the fact that $lim_{xto 0^+} x^x = 1$:
        $$
        begin{align}
        lim_{xto 0^+} frac{x^{x+{1over x}}+ 1}{x^x} &= lim_{xto 0^+} left(x^{x+{1over x}} + 1right) \
        &= 1+ lim_{xto 0^+} x^x cdot x^{1over x} \
        &= 1 + lim_{xto 0^+}x^x cdot lim_{xto 0^+} sqrt[x]{x}\
        &= 1 + 1cdot 0 = 1
        end{align}
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 3 at 14:55









        romanroman

        2,42721226




        2,42721226






























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