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Find $limlimits_{x to 0+} theta (x)$ and $limlimits_{x to 0-} theta (x)$

How to find $limlimits_{ntoinfty} n·(sqrt[n]{a}-1)$?Prove that $limlimits_{xrightarrow+infty}frac{x^k}{a^x} = 0 (a>1,k>0)$Finding $ lbrace a_{n}rbrace $ s.t. $mathop {lim }limits_{n to infty }a_{n}=1$ and $mathop {lim }limits_{n to infty }a_{n}^{n}=2015$Find the limit of $limlimits_{xrightarrow0}frac{x}{tan x}$.$sum limits_{k=1}^{infty} frac{1}{k} cos 2pi k theta$ does not converge as $theta rightarrow 0?$Does $lim frac{a_n}{e^{delta n}}=0$ for every positive $delta$ imply that $lim frac{a_n}{sumlimits_{k=1}^n a_k} =0$Evaluate $limlimits_{xto0} x^p (log(1/x))^m$ where $0<p<1$ and $m>0$, or $p>1 $ and $m>0$Does $limlimits_{xrightarrow infty}f'(x)$ exist?How does one prove $limlimits_{n→+∞} frac{sin(n+1)}{sin(n)}$ does not exist by means of contradiction?How to prove that $limlimits_{nto infty}a_{n+1}=limlimits_{nto infty}a_{n}$

0

$begingroup$

I have went through the Lagrange theorem ($f(b)-f(a)=f'(c)(b-a)$), and now looking at the following problem:

Prove that for positive $x$, the following is true:

$$sqrt{x+1}-sqrt{x} = frac{1}{2sqrt{x+theta (x)}}$$
where $1/4 < theta(x) < 1/2$. Find $limlimits_{x to 0+} theta (x)$ and $limlimits_{x to 0-} theta (x)$.

Struggling with this. I can see that $f(b)-f(a)=x+1-x=1$, which means that $f'(c) = frac{1}{2sqrt{x+theta(x)}}$, but this does not help me much. No integration should be used in this solution.

real-analysis derivatives

share|cite|improve this question

edited Mar 10 at 14:28

rtybase

11.4k31533

asked Mar 10 at 13:44

i squared - Keep it Reali squared - Keep it Real

1,61311027

$endgroup$

add a comment | 

0

$begingroup$

I have went through the Lagrange theorem ($f(b)-f(a)=f'(c)(b-a)$), and now looking at the following problem:

Prove that for positive $x$, the following is true:

$$sqrt{x+1}-sqrt{x} = frac{1}{2sqrt{x+theta (x)}}$$
where $1/4 < theta(x) < 1/2$. Find $limlimits_{x to 0+} theta (x)$ and $limlimits_{x to 0-} theta (x)$.

Struggling with this. I can see that $f(b)-f(a)=x+1-x=1$, which means that $f'(c) = frac{1}{2sqrt{x+theta(x)}}$, but this does not help me much. No integration should be used in this solution.

real-analysis derivatives

share|cite|improve this question

edited Mar 10 at 14:28

rtybase

11.4k31533

asked Mar 10 at 13:44

i squared - Keep it Reali squared - Keep it Real

1,61311027

$endgroup$

add a comment | 

$begingroup$

I have went through the Lagrange theorem ($f(b)-f(a)=f'(c)(b-a)$), and now looking at the following problem:

Prove that for positive $x$, the following is true:

$$sqrt{x+1}-sqrt{x} = frac{1}{2sqrt{x+theta (x)}}$$
where $1/4 < theta(x) < 1/2$. Find $limlimits_{x to 0+} theta (x)$ and $limlimits_{x to 0-} theta (x)$.

Struggling with this. I can see that $f(b)-f(a)=x+1-x=1$, which means that $f'(c) = frac{1}{2sqrt{x+theta(x)}}$, but this does not help me much. No integration should be used in this solution.

real-analysis derivatives

share|cite|improve this question

edited Mar 10 at 14:28

rtybase

11.4k31533

asked Mar 10 at 13:44

i squared - Keep it Reali squared - Keep it Real

1,61311027

$endgroup$

share|cite|improve this question

edited Mar 10 at 14:28

rtybase

11.4k31533

asked Mar 10 at 13:44

i squared - Keep it Reali squared - Keep it Real

1,61311027

share|cite|improve this question

edited Mar 10 at 14:28

rtybase

11.4k31533

asked Mar 10 at 13:44

i squared - Keep it Reali squared - Keep it Real

1,61311027

edited Mar 10 at 14:28

rtybase

11.4k31533

edited Mar 10 at 14:28

rtybase

11.4k31533

asked Mar 10 at 13:44

i squared - Keep it Reali squared - Keep it Real

1,61311027

asked Mar 10 at 13:44

i squared - Keep it Reali squared - Keep it Real

1,61311027

2 Answers
2

active

oldest

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1

$begingroup$

You don't need to use Lagrange's theorem. You can simply use the given relation and solve for $theta(x)$.

$$
theta (x) = frac 14 left(1+ 2 sqrt{x^2+x}-2xright)
$$

The given bounds and the value of the right limit come directly from this expression. The left limit does not make sense as the expression in not defined for negative $x$.

In fact, $theta$ is an increasing function of $x$ defined for $x ge 0$ and you have that
$$
lim_{x to 0^+} theta(x)= frac 14
$$

and

$$
lim_{x to +infty} theta (x) = frac 12.
$$

share|cite|improve this answer

answered Mar 10 at 14:05

PierreCarrePierreCarre

1,480211

$endgroup$

add a comment | 

0

$begingroup$

As a check:

$f(x+1)-f(x)= f'(t)$, where $x <t_x<x+1$.

With $f(x)=√x$ , $x >0$ .

$sqrt{x+1}-√x = (1/2)(t_x)^{-1/2}$, $x <t_x <x+1$.

Comparing with the expression given:

$t_x = x+theta(x)$, we have

$x < x +theta(x) <x+1$,

$0< theta(x) <1$, $x >0$.

Limits $x rightarrow 0^+$, and $x rightarrow infty.$

Solve for $theta (x),$ as done by PierreCarree.

$theta (x) =(1/4)(1+2(sqrt{x^2+x}-2x).$

1)$lim_{x rightarrow 0^+}theta (x)=1/4$.

2)$theta (x)= (1/4)(1+2(sqrt{x^2-x}-x);$

Recall: $(x^2+x) -x^2=$

$(sqrt{x^2+x}+ x)(sqrt{x^2+x}- x);$

$sqrt{x^2+x}-x =dfrac{x}{sqrt{x^2+x}+x}=$

$dfrac{1}{sqrt{1+1/x}+1}.$

Finally:

$lim_{ xrightarrow infty} theta (x)= 1/2.$

share|cite|improve this answer

edited Mar 10 at 18:03

answered Mar 10 at 17:52

Peter SzilasPeter Szilas

11.5k2822

$endgroup$

add a comment | 

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1

$begingroup$

You don't need to use Lagrange's theorem. You can simply use the given relation and solve for $theta(x)$.

$$
theta (x) = frac 14 left(1+ 2 sqrt{x^2+x}-2xright)
$$

The given bounds and the value of the right limit come directly from this expression. The left limit does not make sense as the expression in not defined for negative $x$.

In fact, $theta$ is an increasing function of $x$ defined for $x ge 0$ and you have that
$$
lim_{x to 0^+} theta(x)= frac 14
$$

and

$$
lim_{x to +infty} theta (x) = frac 12.
$$

share|cite|improve this answer

answered Mar 10 at 14:05

PierreCarrePierreCarre

1,480211

$endgroup$

add a comment | 

1

$begingroup$

You don't need to use Lagrange's theorem. You can simply use the given relation and solve for $theta(x)$.

$$
theta (x) = frac 14 left(1+ 2 sqrt{x^2+x}-2xright)
$$

The given bounds and the value of the right limit come directly from this expression. The left limit does not make sense as the expression in not defined for negative $x$.

In fact, $theta$ is an increasing function of $x$ defined for $x ge 0$ and you have that
$$
lim_{x to 0^+} theta(x)= frac 14
$$

and

$$
lim_{x to +infty} theta (x) = frac 12.
$$

share|cite|improve this answer

answered Mar 10 at 14:05

PierreCarrePierreCarre

1,480211

$endgroup$

add a comment | 

$begingroup$

You don't need to use Lagrange's theorem. You can simply use the given relation and solve for $theta(x)$.

$$
theta (x) = frac 14 left(1+ 2 sqrt{x^2+x}-2xright)
$$

The given bounds and the value of the right limit come directly from this expression. The left limit does not make sense as the expression in not defined for negative $x$.

In fact, $theta$ is an increasing function of $x$ defined for $x ge 0$ and you have that
$$
lim_{x to 0^+} theta(x)= frac 14
$$

and

$$
lim_{x to +infty} theta (x) = frac 12.
$$

share|cite|improve this answer

answered Mar 10 at 14:05

PierreCarrePierreCarre

1,480211

$endgroup$

share|cite|improve this answer

answered Mar 10 at 14:05

PierreCarrePierreCarre

1,480211

answered Mar 10 at 14:05

PierreCarrePierreCarre

1,480211

answered Mar 10 at 14:05

PierreCarrePierreCarre

1,480211

0

$begingroup$

As a check:

$f(x+1)-f(x)= f'(t)$, where $x <t_x<x+1$.

With $f(x)=√x$ , $x >0$ .

$sqrt{x+1}-√x = (1/2)(t_x)^{-1/2}$, $x <t_x <x+1$.

Comparing with the expression given:

$t_x = x+theta(x)$, we have

$x < x +theta(x) <x+1$,

$0< theta(x) <1$, $x >0$.

Limits $x rightarrow 0^+$, and $x rightarrow infty.$

Solve for $theta (x),$ as done by PierreCarree.

$theta (x) =(1/4)(1+2(sqrt{x^2+x}-2x).$

1)$lim_{x rightarrow 0^+}theta (x)=1/4$.

2)$theta (x)= (1/4)(1+2(sqrt{x^2-x}-x);$

Recall: $(x^2+x) -x^2=$

$(sqrt{x^2+x}+ x)(sqrt{x^2+x}- x);$

$sqrt{x^2+x}-x =dfrac{x}{sqrt{x^2+x}+x}=$

$dfrac{1}{sqrt{1+1/x}+1}.$

Finally:

$lim_{ xrightarrow infty} theta (x)= 1/2.$

share|cite|improve this answer

edited Mar 10 at 18:03

answered Mar 10 at 17:52

Peter SzilasPeter Szilas

11.5k2822

$endgroup$

add a comment | 

0

$begingroup$

As a check:

$f(x+1)-f(x)= f'(t)$, where $x <t_x<x+1$.

With $f(x)=√x$ , $x >0$ .

$sqrt{x+1}-√x = (1/2)(t_x)^{-1/2}$, $x <t_x <x+1$.

Comparing with the expression given:

$t_x = x+theta(x)$, we have

$x < x +theta(x) <x+1$,

$0< theta(x) <1$, $x >0$.

Limits $x rightarrow 0^+$, and $x rightarrow infty.$

Solve for $theta (x),$ as done by PierreCarree.

$theta (x) =(1/4)(1+2(sqrt{x^2+x}-2x).$

1)$lim_{x rightarrow 0^+}theta (x)=1/4$.

2)$theta (x)= (1/4)(1+2(sqrt{x^2-x}-x);$

Recall: $(x^2+x) -x^2=$

$(sqrt{x^2+x}+ x)(sqrt{x^2+x}- x);$

$sqrt{x^2+x}-x =dfrac{x}{sqrt{x^2+x}+x}=$

$dfrac{1}{sqrt{1+1/x}+1}.$

Finally:

$lim_{ xrightarrow infty} theta (x)= 1/2.$

share|cite|improve this answer

edited Mar 10 at 18:03

answered Mar 10 at 17:52

Peter SzilasPeter Szilas

11.5k2822

$endgroup$

add a comment | 

$begingroup$

As a check:

$f(x+1)-f(x)= f'(t)$, where $x <t_x<x+1$.

With $f(x)=√x$ , $x >0$ .

$sqrt{x+1}-√x = (1/2)(t_x)^{-1/2}$, $x <t_x <x+1$.

Comparing with the expression given:

$t_x = x+theta(x)$, we have

$x < x +theta(x) <x+1$,

$0< theta(x) <1$, $x >0$.

Limits $x rightarrow 0^+$, and $x rightarrow infty.$

Solve for $theta (x),$ as done by PierreCarree.

$theta (x) =(1/4)(1+2(sqrt{x^2+x}-2x).$

1)$lim_{x rightarrow 0^+}theta (x)=1/4$.

2)$theta (x)= (1/4)(1+2(sqrt{x^2-x}-x);$

Recall: $(x^2+x) -x^2=$

$(sqrt{x^2+x}+ x)(sqrt{x^2+x}- x);$

$sqrt{x^2+x}-x =dfrac{x}{sqrt{x^2+x}+x}=$

$dfrac{1}{sqrt{1+1/x}+1}.$

Finally:

$lim_{ xrightarrow infty} theta (x)= 1/2.$

share|cite|improve this answer

edited Mar 10 at 18:03

answered Mar 10 at 17:52

Peter SzilasPeter Szilas

11.5k2822

$endgroup$

share|cite|improve this answer

edited Mar 10 at 18:03

answered Mar 10 at 17:52

Peter SzilasPeter Szilas

11.5k2822

answered Mar 10 at 17:52

Peter SzilasPeter Szilas

11.5k2822

answered Mar 10 at 17:52

Peter SzilasPeter Szilas

11.5k2822