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 $lim_nrightarrow infty n^1/n$
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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 $lim_nrightarrow infty n^1/n$
$begingroup$
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 $f'(1)$ ?
real-analysis
$endgroup$
add a comment |
$begingroup$
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 $f'(1)$ ?
real-analysis
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1
$begingroup$
What is $f$ here?
$endgroup$
– Dr. Sonnhard Graubner
Mar 22 at 14:51
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@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
add a comment |
$begingroup$
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 $f'(1)$ ?
real-analysis
$endgroup$
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
real-analysis
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
add a comment |
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
add a comment |
1 Answer
1
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$begingroup$
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 $f'(1) = m$.
$endgroup$
$begingroup$
thanks u @Umberto
$endgroup$
– jasmine
Mar 22 at 14:53
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
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 $f'(1) = m$.
$endgroup$
$begingroup$
thanks u @Umberto
$endgroup$
– jasmine
Mar 22 at 14:53
add a comment |
$begingroup$
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 $f'(1) = m$.
$endgroup$
$begingroup$
thanks u @Umberto
$endgroup$
– jasmine
Mar 22 at 14:53
add a comment |
$begingroup$
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 $f'(1) = m$.
$endgroup$
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 $f'(1) = m$.
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
add a comment |
$begingroup$
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
add a comment |
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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