Uniformly continuous derivative implies existence of limit












2












$begingroup$



Let $f in C^1([0, +infty))$. Suppose that $lim_{x rightarrow +infty} f(x)=L$ and $f'$ is uniformly continuous.



Show that $$lim_{x rightarrow +infty} f'(x) + f(x)=L$$




I tried to apply L'Hospital's Rule to $frac{e^xf(x)}{e^x}$ since $frac{d}{dx}e^xf(x)=e^x(f'(x)+f(x))$. It seems alright but I didn't use the uniform continuity of $f'$ and it doesn't work for the function $f(x)=frac{sin(x^2)}{x}$ whose derivative is $f'(x)=2cos(x^2)-frac{sin(x^2)}{x^2}$ since $lim_{x rightarrow +infty} f'(x)$ doesn't exist.



Any ideas? Thanks in advance.






real-analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    The L'Hospital trick won't work in cases where the limit of $[f(x) +f'(x)]$ does not exist as in your example.
    $endgroup$
    – RRL
    1 hour ago
















2












$begingroup$



Let $f in C^1([0, +infty))$. Suppose that $lim_{x rightarrow +infty} f(x)=L$ and $f'$ is uniformly continuous.



Show that $$lim_{x rightarrow +infty} f'(x) + f(x)=L$$




I tried to apply L'Hospital's Rule to $frac{e^xf(x)}{e^x}$ since $frac{d}{dx}e^xf(x)=e^x(f'(x)+f(x))$. It seems alright but I didn't use the uniform continuity of $f'$ and it doesn't work for the function $f(x)=frac{sin(x^2)}{x}$ whose derivative is $f'(x)=2cos(x^2)-frac{sin(x^2)}{x^2}$ since $lim_{x rightarrow +infty} f'(x)$ doesn't exist.



Any ideas? Thanks in advance.






real-analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    The L'Hospital trick won't work in cases where the limit of $[f(x) +f'(x)]$ does not exist as in your example.
    $endgroup$
    – RRL
    1 hour ago














2












2








2


1



$begingroup$



Let $f in C^1([0, +infty))$. Suppose that $lim_{x rightarrow +infty} f(x)=L$ and $f'$ is uniformly continuous.



Show that $$lim_{x rightarrow +infty} f'(x) + f(x)=L$$




I tried to apply L'Hospital's Rule to $frac{e^xf(x)}{e^x}$ since $frac{d}{dx}e^xf(x)=e^x(f'(x)+f(x))$. It seems alright but I didn't use the uniform continuity of $f'$ and it doesn't work for the function $f(x)=frac{sin(x^2)}{x}$ whose derivative is $f'(x)=2cos(x^2)-frac{sin(x^2)}{x^2}$ since $lim_{x rightarrow +infty} f'(x)$ doesn't exist.



Any ideas? Thanks in advance.






real-analysis







share|cite|improve this question









$endgroup$





Let $f in C^1([0, +infty))$. Suppose that $lim_{x rightarrow +infty} f(x)=L$ and $f'$ is uniformly continuous.



Show that $$lim_{x rightarrow +infty} f'(x) + f(x)=L$$




I tried to apply L'Hospital's Rule to $frac{e^xf(x)}{e^x}$ since $frac{d}{dx}e^xf(x)=e^x(f'(x)+f(x))$. It seems alright but I didn't use the uniform continuity of $f'$ and it doesn't work for the function $f(x)=frac{sin(x^2)}{x}$ whose derivative is $f'(x)=2cos(x^2)-frac{sin(x^2)}{x^2}$ since $lim_{x rightarrow +infty} f'(x)$ doesn't exist.



Any ideas? Thanks in advance.






real-analysis




real-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 2 hours ago









lzralbulzralbu

697512




697512












  • $begingroup$
    The L'Hospital trick won't work in cases where the limit of $[f(x) +f'(x)]$ does not exist as in your example.
    $endgroup$
    – RRL
    1 hour ago


















  • $begingroup$
    The L'Hospital trick won't work in cases where the limit of $[f(x) +f'(x)]$ does not exist as in your example.
    $endgroup$
    – RRL
    1 hour ago
















$begingroup$
The L'Hospital trick won't work in cases where the limit of $[f(x) +f'(x)]$ does not exist as in your example.
$endgroup$
– RRL
1 hour ago




$begingroup$
The L'Hospital trick won't work in cases where the limit of $[f(x) +f'(x)]$ does not exist as in your example.
$endgroup$
– RRL
1 hour ago










1 Answer
1






active

oldest

votes


















3












$begingroup$

We have $lim_{x to infty} f'(x) = 0$ because,



$$int_0^x f'(t) , dt = f(x) - f(0), \int_0^infty f'(t) , dt = lim_{x to infty}f(x) - f(0) = L - f(0) quad (text{convergent})$$



and $f'$ is uniformly continuous.



To prove this assume that $lim_{x to infty}f'(x) =0$ does not hold and arrive at contradiction with the fact that the integral of $f'$ is convergent.



If $lim_{x to infty} f'(x) = 0$ does not hold then there exists $epsilon_0 > 0$ and a sequence $x_n to infty$ such that $|f'(x_n)| geqslant epsilon_0$ for all $n$. Next apply uniform continuity.



Assume WLOG that $f'(x_n) geqslant epsilon_0$.



There exists by uniform continuity $delta > 0$ such that $|f'(t) - f'(x_n)| < epsilon_0/2 implies f'(t) > epsilon_0/2$ for all $t in [x_n - delta,x_n + delta],$ and



$$ int_{x_n - delta}^{x_n + delta} f'(t) , dt > epsilondelta$$



This violates the Cauchy criterion for convergence of the improper integral since $x_n$ can be arbitrarily large.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I can help you further, but first let me know if these hints makes it obvious to you now.
    $endgroup$
    – RRL
    1 hour ago










  • $begingroup$
    I still can't see how to use uniform continuity. Could you, please, explain it further?
    $endgroup$
    – lzralbu
    1 hour ago










  • $begingroup$
    I shall do so...
    $endgroup$
    – RRL
    1 hour ago










  • $begingroup$
    What about the example given in the question?
    $endgroup$
    – Jens Schwaiger
    48 mins ago










  • $begingroup$
    @JensSchwaiger: $cos(x^2)$ is not uniformly continuous on $[0,infty)$. OP introduced this as a counterexample for the L'Hospital trick. It is not relevant to the actual question where the assumption is that $f'$ is uniformly continuous.
    $endgroup$
    – RRL
    33 mins ago












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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

We have $lim_{x to infty} f'(x) = 0$ because,



$$int_0^x f'(t) , dt = f(x) - f(0), \int_0^infty f'(t) , dt = lim_{x to infty}f(x) - f(0) = L - f(0) quad (text{convergent})$$



and $f'$ is uniformly continuous.



To prove this assume that $lim_{x to infty}f'(x) =0$ does not hold and arrive at contradiction with the fact that the integral of $f'$ is convergent.



If $lim_{x to infty} f'(x) = 0$ does not hold then there exists $epsilon_0 > 0$ and a sequence $x_n to infty$ such that $|f'(x_n)| geqslant epsilon_0$ for all $n$. Next apply uniform continuity.



Assume WLOG that $f'(x_n) geqslant epsilon_0$.



There exists by uniform continuity $delta > 0$ such that $|f'(t) - f'(x_n)| < epsilon_0/2 implies f'(t) > epsilon_0/2$ for all $t in [x_n - delta,x_n + delta],$ and



$$ int_{x_n - delta}^{x_n + delta} f'(t) , dt > epsilondelta$$



This violates the Cauchy criterion for convergence of the improper integral since $x_n$ can be arbitrarily large.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I can help you further, but first let me know if these hints makes it obvious to you now.
    $endgroup$
    – RRL
    1 hour ago










  • $begingroup$
    I still can't see how to use uniform continuity. Could you, please, explain it further?
    $endgroup$
    – lzralbu
    1 hour ago










  • $begingroup$
    I shall do so...
    $endgroup$
    – RRL
    1 hour ago










  • $begingroup$
    What about the example given in the question?
    $endgroup$
    – Jens Schwaiger
    48 mins ago










  • $begingroup$
    @JensSchwaiger: $cos(x^2)$ is not uniformly continuous on $[0,infty)$. OP introduced this as a counterexample for the L'Hospital trick. It is not relevant to the actual question where the assumption is that $f'$ is uniformly continuous.
    $endgroup$
    – RRL
    33 mins ago
















3












$begingroup$

We have $lim_{x to infty} f'(x) = 0$ because,



$$int_0^x f'(t) , dt = f(x) - f(0), \int_0^infty f'(t) , dt = lim_{x to infty}f(x) - f(0) = L - f(0) quad (text{convergent})$$



and $f'$ is uniformly continuous.



To prove this assume that $lim_{x to infty}f'(x) =0$ does not hold and arrive at contradiction with the fact that the integral of $f'$ is convergent.



If $lim_{x to infty} f'(x) = 0$ does not hold then there exists $epsilon_0 > 0$ and a sequence $x_n to infty$ such that $|f'(x_n)| geqslant epsilon_0$ for all $n$. Next apply uniform continuity.



Assume WLOG that $f'(x_n) geqslant epsilon_0$.



There exists by uniform continuity $delta > 0$ such that $|f'(t) - f'(x_n)| < epsilon_0/2 implies f'(t) > epsilon_0/2$ for all $t in [x_n - delta,x_n + delta],$ and



$$ int_{x_n - delta}^{x_n + delta} f'(t) , dt > epsilondelta$$



This violates the Cauchy criterion for convergence of the improper integral since $x_n$ can be arbitrarily large.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I can help you further, but first let me know if these hints makes it obvious to you now.
    $endgroup$
    – RRL
    1 hour ago










  • $begingroup$
    I still can't see how to use uniform continuity. Could you, please, explain it further?
    $endgroup$
    – lzralbu
    1 hour ago










  • $begingroup$
    I shall do so...
    $endgroup$
    – RRL
    1 hour ago










  • $begingroup$
    What about the example given in the question?
    $endgroup$
    – Jens Schwaiger
    48 mins ago










  • $begingroup$
    @JensSchwaiger: $cos(x^2)$ is not uniformly continuous on $[0,infty)$. OP introduced this as a counterexample for the L'Hospital trick. It is not relevant to the actual question where the assumption is that $f'$ is uniformly continuous.
    $endgroup$
    – RRL
    33 mins ago














3












3








3





$begingroup$

We have $lim_{x to infty} f'(x) = 0$ because,



$$int_0^x f'(t) , dt = f(x) - f(0), \int_0^infty f'(t) , dt = lim_{x to infty}f(x) - f(0) = L - f(0) quad (text{convergent})$$



and $f'$ is uniformly continuous.



To prove this assume that $lim_{x to infty}f'(x) =0$ does not hold and arrive at contradiction with the fact that the integral of $f'$ is convergent.



If $lim_{x to infty} f'(x) = 0$ does not hold then there exists $epsilon_0 > 0$ and a sequence $x_n to infty$ such that $|f'(x_n)| geqslant epsilon_0$ for all $n$. Next apply uniform continuity.



Assume WLOG that $f'(x_n) geqslant epsilon_0$.



There exists by uniform continuity $delta > 0$ such that $|f'(t) - f'(x_n)| < epsilon_0/2 implies f'(t) > epsilon_0/2$ for all $t in [x_n - delta,x_n + delta],$ and



$$ int_{x_n - delta}^{x_n + delta} f'(t) , dt > epsilondelta$$



This violates the Cauchy criterion for convergence of the improper integral since $x_n$ can be arbitrarily large.






share|cite|improve this answer











$endgroup$



We have $lim_{x to infty} f'(x) = 0$ because,



$$int_0^x f'(t) , dt = f(x) - f(0), \int_0^infty f'(t) , dt = lim_{x to infty}f(x) - f(0) = L - f(0) quad (text{convergent})$$



and $f'$ is uniformly continuous.



To prove this assume that $lim_{x to infty}f'(x) =0$ does not hold and arrive at contradiction with the fact that the integral of $f'$ is convergent.



If $lim_{x to infty} f'(x) = 0$ does not hold then there exists $epsilon_0 > 0$ and a sequence $x_n to infty$ such that $|f'(x_n)| geqslant epsilon_0$ for all $n$. Next apply uniform continuity.



Assume WLOG that $f'(x_n) geqslant epsilon_0$.



There exists by uniform continuity $delta > 0$ such that $|f'(t) - f'(x_n)| < epsilon_0/2 implies f'(t) > epsilon_0/2$ for all $t in [x_n - delta,x_n + delta],$ and



$$ int_{x_n - delta}^{x_n + delta} f'(t) , dt > epsilondelta$$



This violates the Cauchy criterion for convergence of the improper integral since $x_n$ can be arbitrarily large.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 1 hour ago

























answered 1 hour ago









RRLRRL

54.1k52675




54.1k52675












  • $begingroup$
    I can help you further, but first let me know if these hints makes it obvious to you now.
    $endgroup$
    – RRL
    1 hour ago










  • $begingroup$
    I still can't see how to use uniform continuity. Could you, please, explain it further?
    $endgroup$
    – lzralbu
    1 hour ago










  • $begingroup$
    I shall do so...
    $endgroup$
    – RRL
    1 hour ago










  • $begingroup$
    What about the example given in the question?
    $endgroup$
    – Jens Schwaiger
    48 mins ago










  • $begingroup$
    @JensSchwaiger: $cos(x^2)$ is not uniformly continuous on $[0,infty)$. OP introduced this as a counterexample for the L'Hospital trick. It is not relevant to the actual question where the assumption is that $f'$ is uniformly continuous.
    $endgroup$
    – RRL
    33 mins ago


















  • $begingroup$
    I can help you further, but first let me know if these hints makes it obvious to you now.
    $endgroup$
    – RRL
    1 hour ago










  • $begingroup$
    I still can't see how to use uniform continuity. Could you, please, explain it further?
    $endgroup$
    – lzralbu
    1 hour ago










  • $begingroup$
    I shall do so...
    $endgroup$
    – RRL
    1 hour ago










  • $begingroup$
    What about the example given in the question?
    $endgroup$
    – Jens Schwaiger
    48 mins ago










  • $begingroup$
    @JensSchwaiger: $cos(x^2)$ is not uniformly continuous on $[0,infty)$. OP introduced this as a counterexample for the L'Hospital trick. It is not relevant to the actual question where the assumption is that $f'$ is uniformly continuous.
    $endgroup$
    – RRL
    33 mins ago
















$begingroup$
I can help you further, but first let me know if these hints makes it obvious to you now.
$endgroup$
– RRL
1 hour ago




$begingroup$
I can help you further, but first let me know if these hints makes it obvious to you now.
$endgroup$
– RRL
1 hour ago












$begingroup$
I still can't see how to use uniform continuity. Could you, please, explain it further?
$endgroup$
– lzralbu
1 hour ago




$begingroup$
I still can't see how to use uniform continuity. Could you, please, explain it further?
$endgroup$
– lzralbu
1 hour ago












$begingroup$
I shall do so...
$endgroup$
– RRL
1 hour ago




$begingroup$
I shall do so...
$endgroup$
– RRL
1 hour ago












$begingroup$
What about the example given in the question?
$endgroup$
– Jens Schwaiger
48 mins ago




$begingroup$
What about the example given in the question?
$endgroup$
– Jens Schwaiger
48 mins ago












$begingroup$
@JensSchwaiger: $cos(x^2)$ is not uniformly continuous on $[0,infty)$. OP introduced this as a counterexample for the L'Hospital trick. It is not relevant to the actual question where the assumption is that $f'$ is uniformly continuous.
$endgroup$
– RRL
33 mins ago




$begingroup$
@JensSchwaiger: $cos(x^2)$ is not uniformly continuous on $[0,infty)$. OP introduced this as a counterexample for the L'Hospital trick. It is not relevant to the actual question where the assumption is that $f'$ is uniformly continuous.
$endgroup$
– RRL
33 mins ago


















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