Calculate $lim_xto 0frace^fracln(1+x)x-ex$

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Calculate



$$lim_xto 0frace^fracln(1+x)x-ex$$




My Attempt:



$$lim_xto 0ecdot frace^fracln(1+x)x-1-1fracln(1+x)x-1 cdot fracfracln(1+x)x-1x$$



$$lim_xto 0e.fracfracln(1+x)x-1x$$



I am not able to solve this further. I can easily use L-Hospital to show $$lim_xto 0fracfracln(1+x)x-1x=-frac12$$



How can I prove this without L-Hopital or Taylor expansions?






limits limits-without-lhopital







share|cite|improve this question





















  • One can extend $f(x)=ln(1+x)/x$, which is initially defined on $(-1,infty)$ except for zero, to also be defined at zero and remain continuous, by setting $f(0)=1$. This is the derivative of $e^f(x)$ at zero, i.e. $e^f(0)f'(0)$.
    – Ian
    Sep 1 at 4:39











  • Just an idea: $x-fracx^22<ln(1+x)$ so taking limits gives one-sided inequality I cannot find any suitable quadratic which can be set as an upper bound for $ln(x)$ in some nbd of zero, so that I can apply Sandwich Theorem.
    – Sujit Bhattacharyya
    Sep 1 at 5:04










  • If you forbid Taylor then you need some more analysis. One option is to integrate the inequality $1-t<dfrac11+t<1-t+t^2$ in interval $[0,x]$ and use Squeeze Theorem. Another is put $x=e^t-1$ and evaluate the limit $lim _tto 0dfrac e^t-1-tt^2$ using the definition $e^t=lim_ntoinfty (1+(t/n))^n$.
    – Paramanand Singh
    Sep 1 at 7:13











  • @ParamanandSingh I know the limit $lim_tto0frace^t-1-tt^2=frac12$
    – prog_SAHIL
    Sep 1 at 10:10











  • Then the substitution $x=e^t-1$ solves your problem.
    – Paramanand Singh
    Sep 1 at 11:37














up vote
0
down vote

favorite
1













Calculate



$$lim_xto 0frace^fracln(1+x)x-ex$$




My Attempt:



$$lim_xto 0ecdot frace^fracln(1+x)x-1-1fracln(1+x)x-1 cdot fracfracln(1+x)x-1x$$



$$lim_xto 0e.fracfracln(1+x)x-1x$$



I am not able to solve this further. I can easily use L-Hospital to show $$lim_xto 0fracfracln(1+x)x-1x=-frac12$$



How can I prove this without L-Hopital or Taylor expansions?






limits limits-without-lhopital







share|cite|improve this question





















  • One can extend $f(x)=ln(1+x)/x$, which is initially defined on $(-1,infty)$ except for zero, to also be defined at zero and remain continuous, by setting $f(0)=1$. This is the derivative of $e^f(x)$ at zero, i.e. $e^f(0)f'(0)$.
    – Ian
    Sep 1 at 4:39











  • Just an idea: $x-fracx^22<ln(1+x)$ so taking limits gives one-sided inequality I cannot find any suitable quadratic which can be set as an upper bound for $ln(x)$ in some nbd of zero, so that I can apply Sandwich Theorem.
    – Sujit Bhattacharyya
    Sep 1 at 5:04










  • If you forbid Taylor then you need some more analysis. One option is to integrate the inequality $1-t<dfrac11+t<1-t+t^2$ in interval $[0,x]$ and use Squeeze Theorem. Another is put $x=e^t-1$ and evaluate the limit $lim _tto 0dfrac e^t-1-tt^2$ using the definition $e^t=lim_ntoinfty (1+(t/n))^n$.
    – Paramanand Singh
    Sep 1 at 7:13











  • @ParamanandSingh I know the limit $lim_tto0frace^t-1-tt^2=frac12$
    – prog_SAHIL
    Sep 1 at 10:10











  • Then the substitution $x=e^t-1$ solves your problem.
    – Paramanand Singh
    Sep 1 at 11:37












up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1






Calculate



$$lim_xto 0frace^fracln(1+x)x-ex$$




My Attempt:



$$lim_xto 0ecdot frace^fracln(1+x)x-1-1fracln(1+x)x-1 cdot fracfracln(1+x)x-1x$$



$$lim_xto 0e.fracfracln(1+x)x-1x$$



I am not able to solve this further. I can easily use L-Hospital to show $$lim_xto 0fracfracln(1+x)x-1x=-frac12$$



How can I prove this without L-Hopital or Taylor expansions?






limits limits-without-lhopital







share|cite|improve this question














Calculate



$$lim_xto 0frace^fracln(1+x)x-ex$$




My Attempt:



$$lim_xto 0ecdot frace^fracln(1+x)x-1-1fracln(1+x)x-1 cdot fracfracln(1+x)x-1x$$



$$lim_xto 0e.fracfracln(1+x)x-1x$$



I am not able to solve this further. I can easily use L-Hospital to show $$lim_xto 0fracfracln(1+x)x-1x=-frac12$$



How can I prove this without L-Hopital or Taylor expansions?






limits limits-without-lhopital




limits limits-without-lhopital






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Sep 1 at 3:53









prog_SAHIL

1,471318




1,471318











  • One can extend $f(x)=ln(1+x)/x$, which is initially defined on $(-1,infty)$ except for zero, to also be defined at zero and remain continuous, by setting $f(0)=1$. This is the derivative of $e^f(x)$ at zero, i.e. $e^f(0)f'(0)$.
    – Ian
    Sep 1 at 4:39











  • Just an idea: $x-fracx^22<ln(1+x)$ so taking limits gives one-sided inequality I cannot find any suitable quadratic which can be set as an upper bound for $ln(x)$ in some nbd of zero, so that I can apply Sandwich Theorem.
    – Sujit Bhattacharyya
    Sep 1 at 5:04










  • If you forbid Taylor then you need some more analysis. One option is to integrate the inequality $1-t<dfrac11+t<1-t+t^2$ in interval $[0,x]$ and use Squeeze Theorem. Another is put $x=e^t-1$ and evaluate the limit $lim _tto 0dfrac e^t-1-tt^2$ using the definition $e^t=lim_ntoinfty (1+(t/n))^n$.
    – Paramanand Singh
    Sep 1 at 7:13











  • @ParamanandSingh I know the limit $lim_tto0frace^t-1-tt^2=frac12$
    – prog_SAHIL
    Sep 1 at 10:10











  • Then the substitution $x=e^t-1$ solves your problem.
    – Paramanand Singh
    Sep 1 at 11:37
















  • One can extend $f(x)=ln(1+x)/x$, which is initially defined on $(-1,infty)$ except for zero, to also be defined at zero and remain continuous, by setting $f(0)=1$. This is the derivative of $e^f(x)$ at zero, i.e. $e^f(0)f'(0)$.
    – Ian
    Sep 1 at 4:39











  • Just an idea: $x-fracx^22<ln(1+x)$ so taking limits gives one-sided inequality I cannot find any suitable quadratic which can be set as an upper bound for $ln(x)$ in some nbd of zero, so that I can apply Sandwich Theorem.
    – Sujit Bhattacharyya
    Sep 1 at 5:04










  • If you forbid Taylor then you need some more analysis. One option is to integrate the inequality $1-t<dfrac11+t<1-t+t^2$ in interval $[0,x]$ and use Squeeze Theorem. Another is put $x=e^t-1$ and evaluate the limit $lim _tto 0dfrac e^t-1-tt^2$ using the definition $e^t=lim_ntoinfty (1+(t/n))^n$.
    – Paramanand Singh
    Sep 1 at 7:13











  • @ParamanandSingh I know the limit $lim_tto0frace^t-1-tt^2=frac12$
    – prog_SAHIL
    Sep 1 at 10:10











  • Then the substitution $x=e^t-1$ solves your problem.
    – Paramanand Singh
    Sep 1 at 11:37















One can extend $f(x)=ln(1+x)/x$, which is initially defined on $(-1,infty)$ except for zero, to also be defined at zero and remain continuous, by setting $f(0)=1$. This is the derivative of $e^f(x)$ at zero, i.e. $e^f(0)f'(0)$.
– Ian
Sep 1 at 4:39





One can extend $f(x)=ln(1+x)/x$, which is initially defined on $(-1,infty)$ except for zero, to also be defined at zero and remain continuous, by setting $f(0)=1$. This is the derivative of $e^f(x)$ at zero, i.e. $e^f(0)f'(0)$.
– Ian
Sep 1 at 4:39













Just an idea: $x-fracx^22<ln(1+x)$ so taking limits gives one-sided inequality I cannot find any suitable quadratic which can be set as an upper bound for $ln(x)$ in some nbd of zero, so that I can apply Sandwich Theorem.
– Sujit Bhattacharyya
Sep 1 at 5:04




Just an idea: $x-fracx^22<ln(1+x)$ so taking limits gives one-sided inequality I cannot find any suitable quadratic which can be set as an upper bound for $ln(x)$ in some nbd of zero, so that I can apply Sandwich Theorem.
– Sujit Bhattacharyya
Sep 1 at 5:04












If you forbid Taylor then you need some more analysis. One option is to integrate the inequality $1-t<dfrac11+t<1-t+t^2$ in interval $[0,x]$ and use Squeeze Theorem. Another is put $x=e^t-1$ and evaluate the limit $lim _tto 0dfrac e^t-1-tt^2$ using the definition $e^t=lim_ntoinfty (1+(t/n))^n$.
– Paramanand Singh
Sep 1 at 7:13





If you forbid Taylor then you need some more analysis. One option is to integrate the inequality $1-t<dfrac11+t<1-t+t^2$ in interval $[0,x]$ and use Squeeze Theorem. Another is put $x=e^t-1$ and evaluate the limit $lim _tto 0dfrac e^t-1-tt^2$ using the definition $e^t=lim_ntoinfty (1+(t/n))^n$.
– Paramanand Singh
Sep 1 at 7:13













@ParamanandSingh I know the limit $lim_tto0frace^t-1-tt^2=frac12$
– prog_SAHIL
Sep 1 at 10:10





@ParamanandSingh I know the limit $lim_tto0frace^t-1-tt^2=frac12$
– prog_SAHIL
Sep 1 at 10:10













Then the substitution $x=e^t-1$ solves your problem.
– Paramanand Singh
Sep 1 at 11:37




Then the substitution $x=e^t-1$ solves your problem.
– Paramanand Singh
Sep 1 at 11:37










1 Answer
1






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1
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$$ln(1+x)=x-dfrac12x^2+dfrac13x^3+cdots$$
then
beginalign
e^fracln(1+x)x
&= e^1-frac12x+frac13x^2+cdots \
&= eleft(e^-frac12x+frac13x^2+cdotsright) \
&= eleft(1+left(-frac12x+frac13x^2+cdotsright)+dfrac12left(-frac12x+frac13x^2+cdotsright)^2+cdotsright) \
&= eleft(1-dfrac12x+O(x^2)right)
endalign
then
$$lim_xto 0frace^fracln(1+x)x-ex=lim_xto 0fraceleft(1-dfrac12x+O(x^2)right)-ex=colorblue-dfrac12e$$






share|cite|improve this answer






















  • 'How can I prove this without L-Hopital or Taylor expansions?' – Doesn't this use Taylor expansions?
    – Toby Mak
    Sep 1 at 4:15











  • @TobyMak I don't think this limit can be solved without L-Hopital or Taylor expansions. At least we need $limdfracln(1+x)x$ which needs more calculus and more more details.
    – Nosrati
    Sep 1 at 4:24










  • That's fine, it's still a good method!
    – Toby Mak
    Sep 1 at 4:27










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1 Answer
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up vote
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$$ln(1+x)=x-dfrac12x^2+dfrac13x^3+cdots$$
then
beginalign
e^fracln(1+x)x
&= e^1-frac12x+frac13x^2+cdots \
&= eleft(e^-frac12x+frac13x^2+cdotsright) \
&= eleft(1+left(-frac12x+frac13x^2+cdotsright)+dfrac12left(-frac12x+frac13x^2+cdotsright)^2+cdotsright) \
&= eleft(1-dfrac12x+O(x^2)right)
endalign
then
$$lim_xto 0frace^fracln(1+x)x-ex=lim_xto 0fraceleft(1-dfrac12x+O(x^2)right)-ex=colorblue-dfrac12e$$






share|cite|improve this answer






















  • 'How can I prove this without L-Hopital or Taylor expansions?' – Doesn't this use Taylor expansions?
    – Toby Mak
    Sep 1 at 4:15











  • @TobyMak I don't think this limit can be solved without L-Hopital or Taylor expansions. At least we need $limdfracln(1+x)x$ which needs more calculus and more more details.
    – Nosrati
    Sep 1 at 4:24










  • That's fine, it's still a good method!
    – Toby Mak
    Sep 1 at 4:27














up vote
1
down vote













$$ln(1+x)=x-dfrac12x^2+dfrac13x^3+cdots$$
then
beginalign
e^fracln(1+x)x
&= e^1-frac12x+frac13x^2+cdots \
&= eleft(e^-frac12x+frac13x^2+cdotsright) \
&= eleft(1+left(-frac12x+frac13x^2+cdotsright)+dfrac12left(-frac12x+frac13x^2+cdotsright)^2+cdotsright) \
&= eleft(1-dfrac12x+O(x^2)right)
endalign
then
$$lim_xto 0frace^fracln(1+x)x-ex=lim_xto 0fraceleft(1-dfrac12x+O(x^2)right)-ex=colorblue-dfrac12e$$






share|cite|improve this answer






















  • 'How can I prove this without L-Hopital or Taylor expansions?' – Doesn't this use Taylor expansions?
    – Toby Mak
    Sep 1 at 4:15











  • @TobyMak I don't think this limit can be solved without L-Hopital or Taylor expansions. At least we need $limdfracln(1+x)x$ which needs more calculus and more more details.
    – Nosrati
    Sep 1 at 4:24










  • That's fine, it's still a good method!
    – Toby Mak
    Sep 1 at 4:27












up vote
1
down vote










up vote
1
down vote









$$ln(1+x)=x-dfrac12x^2+dfrac13x^3+cdots$$
then
beginalign
e^fracln(1+x)x
&= e^1-frac12x+frac13x^2+cdots \
&= eleft(e^-frac12x+frac13x^2+cdotsright) \
&= eleft(1+left(-frac12x+frac13x^2+cdotsright)+dfrac12left(-frac12x+frac13x^2+cdotsright)^2+cdotsright) \
&= eleft(1-dfrac12x+O(x^2)right)
endalign
then
$$lim_xto 0frace^fracln(1+x)x-ex=lim_xto 0fraceleft(1-dfrac12x+O(x^2)right)-ex=colorblue-dfrac12e$$






share|cite|improve this answer














$$ln(1+x)=x-dfrac12x^2+dfrac13x^3+cdots$$
then
beginalign
e^fracln(1+x)x
&= e^1-frac12x+frac13x^2+cdots \
&= eleft(e^-frac12x+frac13x^2+cdotsright) \
&= eleft(1+left(-frac12x+frac13x^2+cdotsright)+dfrac12left(-frac12x+frac13x^2+cdotsright)^2+cdotsright) \
&= eleft(1-dfrac12x+O(x^2)right)
endalign
then
$$lim_xto 0frace^fracln(1+x)x-ex=lim_xto 0fraceleft(1-dfrac12x+O(x^2)right)-ex=colorblue-dfrac12e$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Sep 1 at 4:16

























answered Sep 1 at 4:14









Nosrati

22.1k61747




22.1k61747











  • 'How can I prove this without L-Hopital or Taylor expansions?' – Doesn't this use Taylor expansions?
    – Toby Mak
    Sep 1 at 4:15











  • @TobyMak I don't think this limit can be solved without L-Hopital or Taylor expansions. At least we need $limdfracln(1+x)x$ which needs more calculus and more more details.
    – Nosrati
    Sep 1 at 4:24










  • That's fine, it's still a good method!
    – Toby Mak
    Sep 1 at 4:27
















  • 'How can I prove this without L-Hopital or Taylor expansions?' – Doesn't this use Taylor expansions?
    – Toby Mak
    Sep 1 at 4:15











  • @TobyMak I don't think this limit can be solved without L-Hopital or Taylor expansions. At least we need $limdfracln(1+x)x$ which needs more calculus and more more details.
    – Nosrati
    Sep 1 at 4:24










  • That's fine, it's still a good method!
    – Toby Mak
    Sep 1 at 4:27















'How can I prove this without L-Hopital or Taylor expansions?' – Doesn't this use Taylor expansions?
– Toby Mak
Sep 1 at 4:15





'How can I prove this without L-Hopital or Taylor expansions?' – Doesn't this use Taylor expansions?
– Toby Mak
Sep 1 at 4:15













@TobyMak I don't think this limit can be solved without L-Hopital or Taylor expansions. At least we need $limdfracln(1+x)x$ which needs more calculus and more more details.
– Nosrati
Sep 1 at 4:24




@TobyMak I don't think this limit can be solved without L-Hopital or Taylor expansions. At least we need $limdfracln(1+x)x$ which needs more calculus and more more details.
– Nosrati
Sep 1 at 4:24












That's fine, it's still a good method!
– Toby Mak
Sep 1 at 4:27




That's fine, it's still a good method!
– Toby Mak
Sep 1 at 4:27

















 

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