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Proof using derivative information to find limit

limit $lim_x to 49 fracsqrtx-7x-49 .$Can you help me calculate a limit?Trying to find limit by rationalizing numerator with square rootLimit of a functionLimit of real logarithmFind $lim_x rightarrow 1 fracsqrt2x^2-1-1x^3-1$ without using L'Hôpital's ruleHow to calculate this limit $lim_xtoinftyfracsqrt9x^2+2x-3(8x^5-6x+1)^frac 13$Integral proof using comparison theorem

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6

3

$begingroup$

This is the last exercise of a quite challenging exercises paper a friend who is taking calculus has which I'm trying to help. I already helped her doing the other bunch. But this got me. I will appreciate anyone help to see my work and to tell me if is right or If I need to correct something.

The exercise is:

If $f'(a)=1$ for $a>0$, find $lim_x to a fracf(x)-f(a)sqrtx-sqrta$.

What came to my mind was to rationalize the denominator.

$$lim_x to a fracf(x)-f(a)sqrtx-sqrta$$

$$=lim_x to a fracf(x)-f(a)sqrtx-sqrtacdot fracsqrtx+sqrtasqrtx+sqrta$$

$$=lim_x to a frac(f(x)-f(a))(sqrtx+sqrta)x-a$$

$$=lim_x to a left(fracf(x)-f(a)x-acdot (sqrtx+sqrta)right)$$

$$=lim_x to a fracf(x)-f(a)x-acdot lim_x to a(sqrtx+sqrta)$$

$$=f'(a)cdot lim_x to a(sqrtx+sqrta)$$

$$=1cdot lim_x to a(sqrtx+sqrta)$$

$$=lim_x to a(sqrtx+sqrta)$$

$$=sqrta+sqrta$$

$$=2sqrta$$

calculus limits derivatives

share|cite|improve this question

edited 8 hours ago

gi2302

asked 8 hours ago

gi2302gi2302

12077 bronze badges

$endgroup$

add a comment
 | 

6

3

$begingroup$

This is the last exercise of a quite challenging exercises paper a friend who is taking calculus has which I'm trying to help. I already helped her doing the other bunch. But this got me. I will appreciate anyone help to see my work and to tell me if is right or If I need to correct something.

The exercise is:

If $f'(a)=1$ for $a>0$, find $lim_x to a fracf(x)-f(a)sqrtx-sqrta$.

What came to my mind was to rationalize the denominator.

$$lim_x to a fracf(x)-f(a)sqrtx-sqrta$$

$$=lim_x to a fracf(x)-f(a)sqrtx-sqrtacdot fracsqrtx+sqrtasqrtx+sqrta$$

$$=lim_x to a frac(f(x)-f(a))(sqrtx+sqrta)x-a$$

$$=lim_x to a left(fracf(x)-f(a)x-acdot (sqrtx+sqrta)right)$$

$$=lim_x to a fracf(x)-f(a)x-acdot lim_x to a(sqrtx+sqrta)$$

$$=f'(a)cdot lim_x to a(sqrtx+sqrta)$$

$$=1cdot lim_x to a(sqrtx+sqrta)$$

$$=lim_x to a(sqrtx+sqrta)$$

$$=sqrta+sqrta$$

$$=2sqrta$$

calculus limits derivatives

share|cite|improve this question

edited 8 hours ago

gi2302

asked 8 hours ago

gi2302gi2302

12077 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

This is the last exercise of a quite challenging exercises paper a friend who is taking calculus has which I'm trying to help. I already helped her doing the other bunch. But this got me. I will appreciate anyone help to see my work and to tell me if is right or If I need to correct something.

The exercise is:

If $f'(a)=1$ for $a>0$, find $lim_x to a fracf(x)-f(a)sqrtx-sqrta$.

What came to my mind was to rationalize the denominator.

$$lim_x to a fracf(x)-f(a)sqrtx-sqrta$$

$$=lim_x to a fracf(x)-f(a)sqrtx-sqrtacdot fracsqrtx+sqrtasqrtx+sqrta$$

$$=lim_x to a frac(f(x)-f(a))(sqrtx+sqrta)x-a$$

$$=lim_x to a left(fracf(x)-f(a)x-acdot (sqrtx+sqrta)right)$$

$$=lim_x to a fracf(x)-f(a)x-acdot lim_x to a(sqrtx+sqrta)$$

$$=f'(a)cdot lim_x to a(sqrtx+sqrta)$$

$$=1cdot lim_x to a(sqrtx+sqrta)$$

$$=lim_x to a(sqrtx+sqrta)$$

$$=sqrta+sqrta$$

$$=2sqrta$$

calculus limits derivatives

share|cite|improve this question

edited 8 hours ago

gi2302

asked 8 hours ago

gi2302gi2302

12077 bronze badges

$endgroup$

share|cite|improve this question

edited 8 hours ago

gi2302

asked 8 hours ago

gi2302gi2302

12077 bronze badges

share|cite|improve this question

edited 8 hours ago

gi2302

asked 8 hours ago

gi2302gi2302

12077 bronze badges

asked 8 hours ago

gi2302gi2302

12077 bronze badges

asked 8 hours ago

gi2302gi2302

12077 bronze badges

1 Answer
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7

$begingroup$

You made a mistake. (It's not a big one but it impacts your answer).

On Line $3$, you should have had $sqrtx+sqrta$ in numerator.

Now, redo the steps (easy), and you'll end up with $2sqrta$.

share|cite|improve this answer

answered 8 hours ago

Saketh MalyalaSaketh Malyala

11.6k1717 silver badges3939 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  great.. i will fix it.. thanks so much!
  $endgroup$
  – gi2302
  8 hours ago
  

  
  
  
  

add a comment
 | 

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7

$begingroup$

You made a mistake. (It's not a big one but it impacts your answer).

On Line $3$, you should have had $sqrtx+sqrta$ in numerator.

Now, redo the steps (easy), and you'll end up with $2sqrta$.

share|cite|improve this answer

answered 8 hours ago

Saketh MalyalaSaketh Malyala

11.6k1717 silver badges3939 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  great.. i will fix it.. thanks so much!
  $endgroup$
  – gi2302
  8 hours ago
  

  
  
  
  

add a comment
 | 

7

$begingroup$

You made a mistake. (It's not a big one but it impacts your answer).

On Line $3$, you should have had $sqrtx+sqrta$ in numerator.

Now, redo the steps (easy), and you'll end up with $2sqrta$.

share|cite|improve this answer

answered 8 hours ago

Saketh MalyalaSaketh Malyala

11.6k1717 silver badges3939 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  great.. i will fix it.. thanks so much!
  $endgroup$
  – gi2302
  8 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

You made a mistake. (It's not a big one but it impacts your answer).

On Line $3$, you should have had $sqrtx+sqrta$ in numerator.

Now, redo the steps (easy), and you'll end up with $2sqrta$.

share|cite|improve this answer

answered 8 hours ago

Saketh MalyalaSaketh Malyala

11.6k1717 silver badges3939 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Saketh MalyalaSaketh Malyala

11.6k1717 silver badges3939 bronze badges

answered 8 hours ago

Saketh MalyalaSaketh Malyala

11.6k1717 silver badges3939 bronze badges

answered 8 hours ago

Saketh MalyalaSaketh Malyala

11.6k1717 silver badges3939 bronze badges