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How to define a function by continuity [closed]

How to prove uniform continuity?Studying continuity of a functionquestion on limits and their calculationGreen's Function Ode 1Continuity of Popcorn Function (Thomae's Function)How to find this limit : $lim_n to infty frac2-sqrt2+sqrt2 +sqrt2+ cdots n text times4^-n$Continuity of multivariable piecewise function (cos, sin)limit of $a_n = left( frac sqrtk - 1 sqrtk + 1 right )^n, quad kin mathbbN$Check the continuity of a function of two variableshow to define the functions $f(x)$ and $g(x)$ by a given equation

0

$begingroup$

Find y(4), if y(1) = 1.

$$
y = frac2 sqrtx + Ctextln x.
$$

So,
$$
y(1) = frac2 sqrt1 + Ctextln 1quad text???
$$

How can I solve this?

limits analysis

share|cite|improve this question

edited Mar 14 at 14:59

Frits Veerman

7,0312921

asked Mar 14 at 14:55

ElyEly

121

$endgroup$

closed as off-topic by Saad, Thomas Shelby, John Omielan, Xander Henderson, Vinyl_cape_jawa Mar 14 at 20:56

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Thomas Shelby, John Omielan, Xander Henderson, Vinyl_cape_jawa

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Are you sure with $$y(1)=1$$? since we have $$ln(1)=0$$
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 14 at 15:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Exactly, we need to define a function by continuity using limits. I don`t know how to do it(
  $endgroup$
  – Ely
  Mar 14 at 15:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You have a zero downstairs in the limit. Is there any possible value of $C$ which might give you a zero upstairs? If so, this completely determines the function, which should allow you to determine the value of $y(4)$.
  $endgroup$
  – Xander Henderson
  Mar 14 at 19:10
  

  
  
  
  

add a comment | 

0

$begingroup$

Find y(4), if y(1) = 1.

$$
y = frac2 sqrtx + Ctextln x.
$$

So,
$$
y(1) = frac2 sqrt1 + Ctextln 1quad text???
$$

How can I solve this?

limits analysis

share|cite|improve this question

edited Mar 14 at 14:59

Frits Veerman

7,0312921

asked Mar 14 at 14:55

ElyEly

121

$endgroup$

closed as off-topic by Saad, Thomas Shelby, John Omielan, Xander Henderson, Vinyl_cape_jawa Mar 14 at 20:56

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Thomas Shelby, John Omielan, Xander Henderson, Vinyl_cape_jawa

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Are you sure with $$y(1)=1$$? since we have $$ln(1)=0$$
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 14 at 15:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Exactly, we need to define a function by continuity using limits. I don`t know how to do it(
  $endgroup$
  – Ely
  Mar 14 at 15:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You have a zero downstairs in the limit. Is there any possible value of $C$ which might give you a zero upstairs? If so, this completely determines the function, which should allow you to determine the value of $y(4)$.
  $endgroup$
  – Xander Henderson
  Mar 14 at 19:10
  

  
  
  
  

add a comment | 

$begingroup$

Find y(4), if y(1) = 1.

$$
y = frac2 sqrtx + Ctextln x.
$$

So,
$$
y(1) = frac2 sqrt1 + Ctextln 1quad text???
$$

How can I solve this?

limits analysis

share|cite|improve this question

edited Mar 14 at 14:59

Frits Veerman

7,0312921

asked Mar 14 at 14:55

ElyEly

121

$endgroup$

share|cite|improve this question

edited Mar 14 at 14:59

Frits Veerman

7,0312921

asked Mar 14 at 14:55

ElyEly

121

share|cite|improve this question

edited Mar 14 at 14:59

Frits Veerman

7,0312921

asked Mar 14 at 14:55

ElyEly

121

edited Mar 14 at 14:59

Frits Veerman

7,0312921

edited Mar 14 at 14:59

Frits Veerman

7,0312921

asked Mar 14 at 14:55

ElyEly

121

asked Mar 14 at 14:55

ElyEly

121

1 Answer
1

active

oldest

votes

1

$begingroup$

I'm assuming that when you say $y=frac2sqrtx+Cln x$ and $y(1)=1$, you mean that $$ lim_xto1 y(x) = lim_xto 1 frac2sqrtx+Cln x = 1 .$$

Otherwise $y(1)$ is undefined, since $ln(1) = 0$. Solving this above limit you get $C=-2$. Now you should be able to find $y(4)$.

share|cite|improve this answer

answered Mar 14 at 15:42

Dando18Dando18

4,73241235

$endgroup$

add a comment | 

1

$begingroup$

I'm assuming that when you say $y=frac2sqrtx+Cln x$ and $y(1)=1$, you mean that $$ lim_xto1 y(x) = lim_xto 1 frac2sqrtx+Cln x = 1 .$$

Otherwise $y(1)$ is undefined, since $ln(1) = 0$. Solving this above limit you get $C=-2$. Now you should be able to find $y(4)$.

share|cite|improve this answer

answered Mar 14 at 15:42

Dando18Dando18

4,73241235

$endgroup$

add a comment | 

1

$begingroup$

I'm assuming that when you say $y=frac2sqrtx+Cln x$ and $y(1)=1$, you mean that $$ lim_xto1 y(x) = lim_xto 1 frac2sqrtx+Cln x = 1 .$$

Otherwise $y(1)$ is undefined, since $ln(1) = 0$. Solving this above limit you get $C=-2$. Now you should be able to find $y(4)$.

share|cite|improve this answer

answered Mar 14 at 15:42

Dando18Dando18

4,73241235

$endgroup$

add a comment | 

$begingroup$

I'm assuming that when you say $y=frac2sqrtx+Cln x$ and $y(1)=1$, you mean that $$ lim_xto1 y(x) = lim_xto 1 frac2sqrtx+Cln x = 1 .$$

Otherwise $y(1)$ is undefined, since $ln(1) = 0$. Solving this above limit you get $C=-2$. Now you should be able to find $y(4)$.

share|cite|improve this answer

answered Mar 14 at 15:42

Dando18Dando18

4,73241235

$endgroup$

share|cite|improve this answer

answered Mar 14 at 15:42

Dando18Dando18

4,73241235

answered Mar 14 at 15:42

Dando18Dando18

4,73241235

answered Mar 14 at 15:42

Dando18Dando18

4,73241235