$f(x) = ln x + int_0^xsqrt1+sin t,dt$ function

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Let $f$ be a real-valued function defined on the interval $(0,infty)$ by
$$f(x) = ln x + int_0^xsqrt1+sin t,dt.$$

Then prove that there exists $alpha>1$ such that $|f'(x)| < |f(x)|$ for all $x in (alpha,infty )$

$f(x)$ is increasing as $f'(x)>0$, $f'(x)=frac1x+sqrt1+sin x$, after this step not able to proceed.

calculus

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edited Jul 25 at 19:59

egreg

164k1180187

asked Jul 25 at 16:41

Samar Imam Zaidi

1,063316

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up vote
1
down vote

favorite

Let $f$ be a real-valued function defined on the interval $(0,infty)$ by
$$f(x) = ln x + int_0^xsqrt1+sin t,dt.$$

Then prove that there exists $alpha>1$ such that $|f'(x)| < |f(x)|$ for all $x in (alpha,infty )$

$f(x)$ is increasing as $f'(x)>0$, $f'(x)=frac1x+sqrt1+sin x$, after this step not able to proceed.

calculus

share|cite|improve this question

edited Jul 25 at 19:59

egreg

164k1180187

asked Jul 25 at 16:41

Samar Imam Zaidi

1,063316

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Let $f$ be a real-valued function defined on the interval $(0,infty)$ by
$$f(x) = ln x + int_0^xsqrt1+sin t,dt.$$

Then prove that there exists $alpha>1$ such that $|f'(x)| < |f(x)|$ for all $x in (alpha,infty )$

$f(x)$ is increasing as $f'(x)>0$, $f'(x)=frac1x+sqrt1+sin x$, after this step not able to proceed.

calculus

share|cite|improve this question

edited Jul 25 at 19:59

egreg

164k1180187

asked Jul 25 at 16:41

Samar Imam Zaidi

1,063316

Let $f$ be a real-valued function defined on the interval $(0,infty)$ by
$$f(x) = ln x + int_0^xsqrt1+sin t,dt.$$

Then prove that there exists $alpha>1$ such that $|f'(x)| < |f(x)|$ for all $x in (alpha,infty )$

$f(x)$ is increasing as $f'(x)>0$, $f'(x)=frac1x+sqrt1+sin x$, after this step not able to proceed.

calculus

share|cite|improve this question

edited Jul 25 at 19:59

egreg

164k1180187

asked Jul 25 at 16:41

Samar Imam Zaidi

1,063316

share|cite|improve this question

share|cite|improve this question

edited Jul 25 at 19:59

egreg

164k1180187

edited Jul 25 at 19:59

egreg

164k1180187

edited Jul 25 at 19:59

egreg

164k1180187

164k1180187

asked Jul 25 at 16:41

Samar Imam Zaidi

1,063316

asked Jul 25 at 16:41

Samar Imam Zaidi

1,063316

asked Jul 25 at 16:41

Samar Imam Zaidi

1,063316

1,063316

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
0
down vote

Define $g(x)=fracf(x)$ and observe that
$$
lim_xto+infty g(x) = 0.
$$
Therefore, there exists $alpha > 0$ such that for every $x > alpha$, we have $g(x) < 1$.

share|cite|improve this answer

answered Jul 25 at 20:09

Gonzalo Benavides

581317

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you elaborate more on the 3rd line like how did you get $g(x)<1$ ? .
  – Alphanerd
  Jul 26 at 7:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I used the definition of limit with $varepsilon = 1$.
  – Gonzalo Benavides
  Jul 26 at 15:09
  

  
  
  
  

add a comment | 

up vote
0
down vote

Note that for $xgt1$ we have

$$|f'(x)|=left|1over x+sqrt1+sin xright|=1over x+sqrt1+sin xlt1+sqrt2$$

while for $xgt alpha=e^1+sqrt2$ we have $ln xgt1+sqrt2$, in which case

$$|f'(x)|lt1+sqrt2ltln xlt ln x+int_0^xsqrt1+sin tdt=f(x)le|f(x)|$$

share|cite|improve this answer

answered Jul 25 at 20:30

Barry Cipra

56.4k652118

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote

Define $g(x)=fracf(x)$ and observe that
$$
lim_xto+infty g(x) = 0.
$$
Therefore, there exists $alpha > 0$ such that for every $x > alpha$, we have $g(x) < 1$.

share|cite|improve this answer

answered Jul 25 at 20:09

Gonzalo Benavides

581317

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you elaborate more on the 3rd line like how did you get $g(x)<1$ ? .
  – Alphanerd
  Jul 26 at 7:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I used the definition of limit with $varepsilon = 1$.
  – Gonzalo Benavides
  Jul 26 at 15:09
  

  
  
  
  

add a comment | 

up vote
0
down vote

Define $g(x)=fracf(x)$ and observe that
$$
lim_xto+infty g(x) = 0.
$$
Therefore, there exists $alpha > 0$ such that for every $x > alpha$, we have $g(x) < 1$.

share|cite|improve this answer

answered Jul 25 at 20:09

Gonzalo Benavides

581317

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you elaborate more on the 3rd line like how did you get $g(x)<1$ ? .
  – Alphanerd
  Jul 26 at 7:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I used the definition of limit with $varepsilon = 1$.
  – Gonzalo Benavides
  Jul 26 at 15:09
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

Define $g(x)=fracf(x)$ and observe that
$$
lim_xto+infty g(x) = 0.
$$
Therefore, there exists $alpha > 0$ such that for every $x > alpha$, we have $g(x) < 1$.

share|cite|improve this answer

answered Jul 25 at 20:09

Gonzalo Benavides

581317

Define $g(x)=fracf(x)$ and observe that
$$
lim_xto+infty g(x) = 0.
$$
Therefore, there exists $alpha > 0$ such that for every $x > alpha$, we have $g(x) < 1$.

share|cite|improve this answer

answered Jul 25 at 20:09

Gonzalo Benavides

581317

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 20:09

Gonzalo Benavides

581317

answered Jul 25 at 20:09

Gonzalo Benavides

581317

answered Jul 25 at 20:09

Gonzalo Benavides

581317

581317

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you elaborate more on the 3rd line like how did you get $g(x)<1$ ? .
  – Alphanerd
  Jul 26 at 7:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I used the definition of limit with $varepsilon = 1$.
  – Gonzalo Benavides
  Jul 26 at 15:09
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could you elaborate more on the 3rd line like how did you get $g(x)<1$ ? .
  – Alphanerd
  Jul 26 at 7:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I used the definition of limit with $varepsilon = 1$.
  – Gonzalo Benavides
  Jul 26 at 15:09
  

  
  
  
  

Could you elaborate more on the 3rd line like how did you get $g(x)<1$ ? .
– Alphanerd
Jul 26 at 7:00

Could you elaborate more on the 3rd line like how did you get $g(x)<1$ ? .
– Alphanerd
Jul 26 at 7:00

I used the definition of limit with $varepsilon = 1$.
– Gonzalo Benavides
Jul 26 at 15:09

I used the definition of limit with $varepsilon = 1$.
– Gonzalo Benavides
Jul 26 at 15:09

add a comment | 

up vote
0
down vote

Note that for $xgt1$ we have

$$|f'(x)|=left|1over x+sqrt1+sin xright|=1over x+sqrt1+sin xlt1+sqrt2$$

while for $xgt alpha=e^1+sqrt2$ we have $ln xgt1+sqrt2$, in which case

$$|f'(x)|lt1+sqrt2ltln xlt ln x+int_0^xsqrt1+sin tdt=f(x)le|f(x)|$$

share|cite|improve this answer

answered Jul 25 at 20:30

Barry Cipra

56.4k652118

add a comment | 

up vote
0
down vote

Note that for $xgt1$ we have

$$|f'(x)|=left|1over x+sqrt1+sin xright|=1over x+sqrt1+sin xlt1+sqrt2$$

while for $xgt alpha=e^1+sqrt2$ we have $ln xgt1+sqrt2$, in which case

$$|f'(x)|lt1+sqrt2ltln xlt ln x+int_0^xsqrt1+sin tdt=f(x)le|f(x)|$$

share|cite|improve this answer

answered Jul 25 at 20:30

Barry Cipra

56.4k652118

add a comment | 

up vote
0
down vote

up vote
0
down vote

Note that for $xgt1$ we have

$$|f'(x)|=left|1over x+sqrt1+sin xright|=1over x+sqrt1+sin xlt1+sqrt2$$

while for $xgt alpha=e^1+sqrt2$ we have $ln xgt1+sqrt2$, in which case

$$|f'(x)|lt1+sqrt2ltln xlt ln x+int_0^xsqrt1+sin tdt=f(x)le|f(x)|$$

share|cite|improve this answer

answered Jul 25 at 20:30

Barry Cipra

56.4k652118

Note that for $xgt1$ we have

$$|f'(x)|=left|1over x+sqrt1+sin xright|=1over x+sqrt1+sin xlt1+sqrt2$$

while for $xgt alpha=e^1+sqrt2$ we have $ln xgt1+sqrt2$, in which case

$$|f'(x)|lt1+sqrt2ltln xlt ln x+int_0^xsqrt1+sin tdt=f(x)le|f(x)|$$

share|cite|improve this answer

answered Jul 25 at 20:30

Barry Cipra

56.4k652118

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 20:30

Barry Cipra

56.4k652118

answered Jul 25 at 20:30

Barry Cipra

56.4k652118

answered Jul 25 at 20:30

Barry Cipra

56.4k652118

56.4k652118

add a comment | 

add a comment | 

 

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