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what is the value of $f'(1)$?


$lim_x to +infty fracln(x)x$ using Taylor seriesWhat is the limsup of the following sequence of sets?Find the value of $lambda$Prove that exists $lim_xrightarrow infty fracf(x)x$ and determine its valuefind the value of $lim_n rightarrow inftyfrac1s_nsum_i=1^na_ix_i$What is the limit of the following sequencehow to find the value of $alpha $ the given series is converge ..Show the series $sum fracx^n(2n)!$ is covergent for all $xinmathbb R$How can I find $lim_nrightarrowinfty(1+fracxn)^sqrtn$?Using the logarithm, find $lim_nrightarrow infty n^1/n$













1












$begingroup$


what is the value of $f'(1)$ ?



$lim_n rightarrow infty frac(1+ 1/n)^m -1 1+ 1/n -1$



i know that $f'(1)=frac(1+ 1/n)^m -1 1+ 1/n -1=frac(1+ 1/n)^m -1 1/n $



but here im confused how can i find the value of $f'(1)$ ?










share|cite|improve this question









$endgroup$







  • 1




    $begingroup$
    What is $f$ here?
    $endgroup$
    – Dr. Sonnhard Graubner
    Mar 22 at 14:51










  • $begingroup$
    @Dr.SonnhardGraubner see umberto answer
    $endgroup$
    – jasmine
    Mar 22 at 14:54






  • 1




    $begingroup$
    So is this the answer for which you looking for?
    $endgroup$
    – Dr. Sonnhard Graubner
    Mar 22 at 14:55










  • $begingroup$
    ya,,,,exactly @Dr.SonnhardGraubner
    $endgroup$
    – jasmine
    Mar 22 at 14:56















1












$begingroup$


what is the value of $f'(1)$ ?



$lim_n rightarrow infty frac(1+ 1/n)^m -1 1+ 1/n -1$



i know that $f'(1)=frac(1+ 1/n)^m -1 1+ 1/n -1=frac(1+ 1/n)^m -1 1/n $



but here im confused how can i find the value of $f'(1)$ ?










share|cite|improve this question









$endgroup$







  • 1




    $begingroup$
    What is $f$ here?
    $endgroup$
    – Dr. Sonnhard Graubner
    Mar 22 at 14:51










  • $begingroup$
    @Dr.SonnhardGraubner see umberto answer
    $endgroup$
    – jasmine
    Mar 22 at 14:54






  • 1




    $begingroup$
    So is this the answer for which you looking for?
    $endgroup$
    – Dr. Sonnhard Graubner
    Mar 22 at 14:55










  • $begingroup$
    ya,,,,exactly @Dr.SonnhardGraubner
    $endgroup$
    – jasmine
    Mar 22 at 14:56













1












1








1





$begingroup$


what is the value of $f'(1)$ ?



$lim_n rightarrow infty frac(1+ 1/n)^m -1 1+ 1/n -1$



i know that $f'(1)=frac(1+ 1/n)^m -1 1+ 1/n -1=frac(1+ 1/n)^m -1 1/n $



but here im confused how can i find the value of $f'(1)$ ?










share|cite|improve this question









$endgroup$




what is the value of $f'(1)$ ?



$lim_n rightarrow infty frac(1+ 1/n)^m -1 1+ 1/n -1$



i know that $f'(1)=frac(1+ 1/n)^m -1 1+ 1/n -1=frac(1+ 1/n)^m -1 1/n $



but here im confused how can i find the value of $f'(1)$ ?







real-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 22 at 14:48









jasminejasmine

1,955420




1,955420







  • 1




    $begingroup$
    What is $f$ here?
    $endgroup$
    – Dr. Sonnhard Graubner
    Mar 22 at 14:51










  • $begingroup$
    @Dr.SonnhardGraubner see umberto answer
    $endgroup$
    – jasmine
    Mar 22 at 14:54






  • 1




    $begingroup$
    So is this the answer for which you looking for?
    $endgroup$
    – Dr. Sonnhard Graubner
    Mar 22 at 14:55










  • $begingroup$
    ya,,,,exactly @Dr.SonnhardGraubner
    $endgroup$
    – jasmine
    Mar 22 at 14:56












  • 1




    $begingroup$
    What is $f$ here?
    $endgroup$
    – Dr. Sonnhard Graubner
    Mar 22 at 14:51










  • $begingroup$
    @Dr.SonnhardGraubner see umberto answer
    $endgroup$
    – jasmine
    Mar 22 at 14:54






  • 1




    $begingroup$
    So is this the answer for which you looking for?
    $endgroup$
    – Dr. Sonnhard Graubner
    Mar 22 at 14:55










  • $begingroup$
    ya,,,,exactly @Dr.SonnhardGraubner
    $endgroup$
    – jasmine
    Mar 22 at 14:56







1




1




$begingroup$
What is $f$ here?
$endgroup$
– Dr. Sonnhard Graubner
Mar 22 at 14:51




$begingroup$
What is $f$ here?
$endgroup$
– Dr. Sonnhard Graubner
Mar 22 at 14:51












$begingroup$
@Dr.SonnhardGraubner see umberto answer
$endgroup$
– jasmine
Mar 22 at 14:54




$begingroup$
@Dr.SonnhardGraubner see umberto answer
$endgroup$
– jasmine
Mar 22 at 14:54




1




1




$begingroup$
So is this the answer for which you looking for?
$endgroup$
– Dr. Sonnhard Graubner
Mar 22 at 14:55




$begingroup$
So is this the answer for which you looking for?
$endgroup$
– Dr. Sonnhard Graubner
Mar 22 at 14:55












$begingroup$
ya,,,,exactly @Dr.SonnhardGraubner
$endgroup$
– jasmine
Mar 22 at 14:56




$begingroup$
ya,,,,exactly @Dr.SonnhardGraubner
$endgroup$
– jasmine
Mar 22 at 14:56










1 Answer
1






active

oldest

votes


















4












$begingroup$

It looks like $f(x) = x^m$ and you are evaluating the limit $$f'(x) = lim_x to 1fracf(x) - f(1)x-1.$$
Since $f'(x) = mx^m-1$ (which is hopefully known) you get $f'(1) = m$.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    thanks u @Umberto
    $endgroup$
    – jasmine
    Mar 22 at 14:53











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes









4












$begingroup$

It looks like $f(x) = x^m$ and you are evaluating the limit $$f'(x) = lim_x to 1fracf(x) - f(1)x-1.$$
Since $f'(x) = mx^m-1$ (which is hopefully known) you get $f'(1) = m$.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    thanks u @Umberto
    $endgroup$
    – jasmine
    Mar 22 at 14:53















4












$begingroup$

It looks like $f(x) = x^m$ and you are evaluating the limit $$f'(x) = lim_x to 1fracf(x) - f(1)x-1.$$
Since $f'(x) = mx^m-1$ (which is hopefully known) you get $f'(1) = m$.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    thanks u @Umberto
    $endgroup$
    – jasmine
    Mar 22 at 14:53













4












4








4





$begingroup$

It looks like $f(x) = x^m$ and you are evaluating the limit $$f'(x) = lim_x to 1fracf(x) - f(1)x-1.$$
Since $f'(x) = mx^m-1$ (which is hopefully known) you get $f'(1) = m$.






share|cite|improve this answer









$endgroup$



It looks like $f(x) = x^m$ and you are evaluating the limit $$f'(x) = lim_x to 1fracf(x) - f(1)x-1.$$
Since $f'(x) = mx^m-1$ (which is hopefully known) you get $f'(1) = m$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Mar 22 at 14:52









Umberto P.Umberto P.

40.3k13370




40.3k13370











  • $begingroup$
    thanks u @Umberto
    $endgroup$
    – jasmine
    Mar 22 at 14:53
















  • $begingroup$
    thanks u @Umberto
    $endgroup$
    – jasmine
    Mar 22 at 14:53















$begingroup$
thanks u @Umberto
$endgroup$
– jasmine
Mar 22 at 14:53




$begingroup$
thanks u @Umberto
$endgroup$
– jasmine
Mar 22 at 14:53

















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