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












1














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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  • 1




    Is $x$ meant to tend to $0$ rather than $infty$? Maybe $0^+$?
    – Olivier Moschetta
    yesterday












  • I believe it should be $xto 0^+$ as pointed out by @OlivierMoschetta, otherwise the limit does not exist
    – roman
    yesterday










  • Yea it's 0. Sorry for inconvenience.
    – Mathsaddict
    yesterday
















1














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




















  • 1




    Is $x$ meant to tend to $0$ rather than $infty$? Maybe $0^+$?
    – Olivier Moschetta
    yesterday












  • I believe it should be $xto 0^+$ as pointed out by @OlivierMoschetta, otherwise the limit does not exist
    – roman
    yesterday










  • Yea it's 0. Sorry for inconvenience.
    – Mathsaddict
    yesterday














1












1








1


2





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















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 yesterday

























asked yesterday









Mathsaddict

2608




2608








  • 1




    Is $x$ meant to tend to $0$ rather than $infty$? Maybe $0^+$?
    – Olivier Moschetta
    yesterday












  • I believe it should be $xto 0^+$ as pointed out by @OlivierMoschetta, otherwise the limit does not exist
    – roman
    yesterday










  • Yea it's 0. Sorry for inconvenience.
    – Mathsaddict
    yesterday














  • 1




    Is $x$ meant to tend to $0$ rather than $infty$? Maybe $0^+$?
    – Olivier Moschetta
    yesterday












  • I believe it should be $xto 0^+$ as pointed out by @OlivierMoschetta, otherwise the limit does not exist
    – roman
    yesterday










  • Yea it's 0. Sorry for inconvenience.
    – Mathsaddict
    yesterday








1




1




Is $x$ meant to tend to $0$ rather than $infty$? Maybe $0^+$?
– Olivier Moschetta
yesterday






Is $x$ meant to tend to $0$ rather than $infty$? Maybe $0^+$?
– Olivier Moschetta
yesterday














I believe it should be $xto 0^+$ as pointed out by @OlivierMoschetta, otherwise the limit does not exist
– roman
yesterday




I believe it should be $xto 0^+$ as pointed out by @OlivierMoschetta, otherwise the limit does not exist
– roman
yesterday












Yea it's 0. Sorry for inconvenience.
– Mathsaddict
yesterday




Yea it's 0. Sorry for inconvenience.
– Mathsaddict
yesterday










2 Answers
2






active

oldest

votes


















2














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





















  • 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.
    – Mathsaddict
    yesterday










  • 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 $|f(x)|^{g(x)}leq |f(x)|rightarrow 0$.
    – Olivier Moschetta
    yesterday



















2














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



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
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    2 Answers
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    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    2














    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





















    • 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.
      – Mathsaddict
      yesterday










    • 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 $|f(x)|^{g(x)}leq |f(x)|rightarrow 0$.
      – Olivier Moschetta
      yesterday
















    2














    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





















    • 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.
      – Mathsaddict
      yesterday










    • 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 $|f(x)|^{g(x)}leq |f(x)|rightarrow 0$.
      – Olivier Moschetta
      yesterday














    2












    2








    2






    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












    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 yesterday









    Olivier Moschetta

    2,8111411




    2,8111411












    • 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.
      – Mathsaddict
      yesterday










    • 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 $|f(x)|^{g(x)}leq |f(x)|rightarrow 0$.
      – Olivier Moschetta
      yesterday


















    • 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.
      – Mathsaddict
      yesterday










    • 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 $|f(x)|^{g(x)}leq |f(x)|rightarrow 0$.
      – Olivier Moschetta
      yesterday
















    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.
    – Mathsaddict
    yesterday




    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.
    – Mathsaddict
    yesterday












    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 $|f(x)|^{g(x)}leq |f(x)|rightarrow 0$.
    – Olivier Moschetta
    yesterday




    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 $|f(x)|^{g(x)}leq |f(x)|rightarrow 0$.
    – Olivier Moschetta
    yesterday











    2














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



    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


























      2














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



      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
























        2












        2








        2






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



        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












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



        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 yesterday









        roman

        1,97121221




        1,97121221






























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