Find the limit of $lim_{ xto 14} frac{x^2-14x}{x^2-196}$

1

$begingroup$

Find the following limit when $$lim_{ xto 14} frac{x^2-14x}{x^2-196}$$ I have find that subbing in 14 gives a null answer. But am having trouble factorizing this equation

calculus

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edited Apr 10 '15 at 4:49

RE60K

14k22155

asked Apr 10 '15 at 4:46

TinaTina

91

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Try L'Hôpital's rule
  $endgroup$
  – Kevin
  Apr 10 '15 at 4:49
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Kevin it's too overboard.
  $endgroup$
  – RE60K
  Apr 10 '15 at 4:49
  

  
  
  
  

add a comment | 

1

$begingroup$

Find the following limit when $$lim_{ xto 14} frac{x^2-14x}{x^2-196}$$ I have find that subbing in 14 gives a null answer. But am having trouble factorizing this equation

calculus

share|cite|improve this question

edited Apr 10 '15 at 4:49

RE60K

14k22155

asked Apr 10 '15 at 4:46

TinaTina

91

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Try L'Hôpital's rule
  $endgroup$
  – Kevin
  Apr 10 '15 at 4:49
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Kevin it's too overboard.
  $endgroup$
  – RE60K
  Apr 10 '15 at 4:49
  

  
  
  
  

add a comment | 

1

1

1

$begingroup$

Find the following limit when $$lim_{ xto 14} frac{x^2-14x}{x^2-196}$$ I have find that subbing in 14 gives a null answer. But am having trouble factorizing this equation

calculus

share|cite|improve this question

edited Apr 10 '15 at 4:49

RE60K

14k22155

asked Apr 10 '15 at 4:46

TinaTina

91

$endgroup$

Find the following limit when $$lim_{ xto 14} frac{x^2-14x}{x^2-196}$$ I have find that subbing in 14 gives a null answer. But am having trouble factorizing this equation

calculus

calculus

share|cite|improve this question

edited Apr 10 '15 at 4:49

RE60K

14k22155

asked Apr 10 '15 at 4:46

TinaTina

91

share|cite|improve this question

edited Apr 10 '15 at 4:49

RE60K

14k22155

asked Apr 10 '15 at 4:46

TinaTina

91

share|cite|improve this question

share|cite|improve this question

edited Apr 10 '15 at 4:49

RE60K

14k22155

edited Apr 10 '15 at 4:49

RE60K

14k22155

edited Apr 10 '15 at 4:49

RE60K

14k22155

14k22155

asked Apr 10 '15 at 4:46

TinaTina

91

asked Apr 10 '15 at 4:46

TinaTina

91

asked Apr 10 '15 at 4:46

TinaTina

91

91

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Try L'Hôpital's rule
  $endgroup$
  – Kevin
  Apr 10 '15 at 4:49
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Kevin it's too overboard.
  $endgroup$
  – RE60K
  Apr 10 '15 at 4:49
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Try L'Hôpital's rule
  $endgroup$
  – Kevin
  Apr 10 '15 at 4:49
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Kevin it's too overboard.
  $endgroup$
  – RE60K
  Apr 10 '15 at 4:49
  

  
  
  
  

$begingroup$
Try L'Hôpital's rule
$endgroup$
– Kevin
Apr 10 '15 at 4:49

$begingroup$
Try L'Hôpital's rule
$endgroup$
– Kevin
Apr 10 '15 at 4:49

2

2

$begingroup$
@Kevin it's too overboard.
$endgroup$
– RE60K
Apr 10 '15 at 4:49

$begingroup$
@Kevin it's too overboard.
$endgroup$
– RE60K
Apr 10 '15 at 4:49

add a comment | 

5 Answers
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4

$begingroup$

$$dfrac{x^2-14x}{x^2-196}=dfrac{x(x-14)}{(x-14)(x+14)}$$

share|cite|improve this answer

answered Apr 10 '15 at 4:50

BumblebeeBumblebee

9,74612551

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2

$begingroup$

$$frac{x^2-14x}{x^2-196}=frac{x(x-14)}{(x+14)(x-14)}={xover x+14}$$ Then you will find the limit$={14over28}={1over2}$.

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edited Jan 22 at 3:42

Math Lover

15910

answered Apr 10 '15 at 4:51

FreyFrey

648312

$endgroup$

add a comment | 

1

$begingroup$

$$x^2-14x=x(x-14)text{ and } x^2-196=x^2-14^2=(x-14)(x+14)$$

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answered Apr 10 '15 at 4:50

RE60KRE60K

14k22155

$endgroup$

add a comment | 

1

$begingroup$

Since x tends to 14 and when you replace x with 14 you get 0/0 condition. Apply L'hospital rule i.e. differentiate both numerator and denominator w.r.t. x. Apply the limit and you have your answer.

share|cite|improve this answer

answered Apr 10 '15 at 7:44

AjayAjay

897

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0

$begingroup$

Hint: when $x neq 14$ we have
$$
frac{x^2-14x}{x^2-196} = frac{x(x-14)}{(x+14)(x-14)} = frac{x}{x+14}.
$$

share|cite|improve this answer

answered Apr 10 '15 at 4:50

Yuval FilmusYuval Filmus

48.7k472145

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5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

active

oldest

votes

active

oldest

votes

4

$begingroup$

$$dfrac{x^2-14x}{x^2-196}=dfrac{x(x-14)}{(x-14)(x+14)}$$

share|cite|improve this answer

answered Apr 10 '15 at 4:50

BumblebeeBumblebee

9,74612551

$endgroup$

add a comment | 

4

$begingroup$

$$dfrac{x^2-14x}{x^2-196}=dfrac{x(x-14)}{(x-14)(x+14)}$$

share|cite|improve this answer

answered Apr 10 '15 at 4:50

BumblebeeBumblebee

9,74612551

$endgroup$

add a comment | 

4

4

4

$begingroup$

$$dfrac{x^2-14x}{x^2-196}=dfrac{x(x-14)}{(x-14)(x+14)}$$

share|cite|improve this answer

answered Apr 10 '15 at 4:50

BumblebeeBumblebee

9,74612551

$endgroup$

$$dfrac{x^2-14x}{x^2-196}=dfrac{x(x-14)}{(x-14)(x+14)}$$

share|cite|improve this answer

answered Apr 10 '15 at 4:50

BumblebeeBumblebee

9,74612551

share|cite|improve this answer

share|cite|improve this answer

answered Apr 10 '15 at 4:50

BumblebeeBumblebee

9,74612551

answered Apr 10 '15 at 4:50

BumblebeeBumblebee

9,74612551

answered Apr 10 '15 at 4:50

BumblebeeBumblebee

9,74612551

9,74612551

add a comment | 

add a comment | 

2

$begingroup$

$$frac{x^2-14x}{x^2-196}=frac{x(x-14)}{(x+14)(x-14)}={xover x+14}$$ Then you will find the limit$={14over28}={1over2}$.

share|cite|improve this answer

edited Jan 22 at 3:42

Math Lover

15910

answered Apr 10 '15 at 4:51

FreyFrey

648312

$endgroup$

add a comment | 

2

$begingroup$

$$frac{x^2-14x}{x^2-196}=frac{x(x-14)}{(x+14)(x-14)}={xover x+14}$$ Then you will find the limit$={14over28}={1over2}$.

share|cite|improve this answer

edited Jan 22 at 3:42

Math Lover

15910

answered Apr 10 '15 at 4:51

FreyFrey

648312

$endgroup$

add a comment | 

2

2

2

$begingroup$

$$frac{x^2-14x}{x^2-196}=frac{x(x-14)}{(x+14)(x-14)}={xover x+14}$$ Then you will find the limit$={14over28}={1over2}$.

share|cite|improve this answer

edited Jan 22 at 3:42

Math Lover

15910

answered Apr 10 '15 at 4:51

FreyFrey

648312

$endgroup$

$$frac{x^2-14x}{x^2-196}=frac{x(x-14)}{(x+14)(x-14)}={xover x+14}$$ Then you will find the limit$={14over28}={1over2}$.

share|cite|improve this answer

edited Jan 22 at 3:42

Math Lover

15910

answered Apr 10 '15 at 4:51

FreyFrey

648312

share|cite|improve this answer

share|cite|improve this answer

edited Jan 22 at 3:42

Math Lover

15910

edited Jan 22 at 3:42

Math Lover

15910

edited Jan 22 at 3:42

Math Lover

15910

15910

answered Apr 10 '15 at 4:51

FreyFrey

648312

answered Apr 10 '15 at 4:51

FreyFrey

648312

answered Apr 10 '15 at 4:51

FreyFrey

648312

648312

add a comment | 

add a comment | 

1

$begingroup$

$$x^2-14x=x(x-14)text{ and } x^2-196=x^2-14^2=(x-14)(x+14)$$

share|cite|improve this answer

answered Apr 10 '15 at 4:50

RE60KRE60K

14k22155

$endgroup$

add a comment | 

1

$begingroup$

$$x^2-14x=x(x-14)text{ and } x^2-196=x^2-14^2=(x-14)(x+14)$$

share|cite|improve this answer

answered Apr 10 '15 at 4:50

RE60KRE60K

14k22155

$endgroup$

add a comment | 

1

1

1

$begingroup$

$$x^2-14x=x(x-14)text{ and } x^2-196=x^2-14^2=(x-14)(x+14)$$

share|cite|improve this answer

answered Apr 10 '15 at 4:50

RE60KRE60K

14k22155

$endgroup$

$$x^2-14x=x(x-14)text{ and } x^2-196=x^2-14^2=(x-14)(x+14)$$

share|cite|improve this answer

answered Apr 10 '15 at 4:50

RE60KRE60K

14k22155

share|cite|improve this answer

share|cite|improve this answer

answered Apr 10 '15 at 4:50

RE60KRE60K

14k22155

answered Apr 10 '15 at 4:50

RE60KRE60K

14k22155

answered Apr 10 '15 at 4:50

RE60KRE60K

14k22155

14k22155

add a comment | 

add a comment | 

1

$begingroup$

Since x tends to 14 and when you replace x with 14 you get 0/0 condition. Apply L'hospital rule i.e. differentiate both numerator and denominator w.r.t. x. Apply the limit and you have your answer.

share|cite|improve this answer

answered Apr 10 '15 at 7:44

AjayAjay

897

$endgroup$

add a comment | 

1

$begingroup$

Since x tends to 14 and when you replace x with 14 you get 0/0 condition. Apply L'hospital rule i.e. differentiate both numerator and denominator w.r.t. x. Apply the limit and you have your answer.

share|cite|improve this answer

answered Apr 10 '15 at 7:44

AjayAjay

897

$endgroup$

add a comment | 

1

1

1

$begingroup$

Since x tends to 14 and when you replace x with 14 you get 0/0 condition. Apply L'hospital rule i.e. differentiate both numerator and denominator w.r.t. x. Apply the limit and you have your answer.

share|cite|improve this answer

answered Apr 10 '15 at 7:44

AjayAjay

897

$endgroup$

Since x tends to 14 and when you replace x with 14 you get 0/0 condition. Apply L'hospital rule i.e. differentiate both numerator and denominator w.r.t. x. Apply the limit and you have your answer.

share|cite|improve this answer

answered Apr 10 '15 at 7:44

AjayAjay

897

share|cite|improve this answer

share|cite|improve this answer

answered Apr 10 '15 at 7:44

AjayAjay

897

answered Apr 10 '15 at 7:44

AjayAjay

897

answered Apr 10 '15 at 7:44

AjayAjay

897

897

add a comment | 

add a comment | 

0

$begingroup$

Hint: when $x neq 14$ we have
$$
frac{x^2-14x}{x^2-196} = frac{x(x-14)}{(x+14)(x-14)} = frac{x}{x+14}.
$$

share|cite|improve this answer

answered Apr 10 '15 at 4:50

Yuval FilmusYuval Filmus

48.7k472145

$endgroup$

add a comment | 

0

$begingroup$

Hint: when $x neq 14$ we have
$$
frac{x^2-14x}{x^2-196} = frac{x(x-14)}{(x+14)(x-14)} = frac{x}{x+14}.
$$

share|cite|improve this answer

answered Apr 10 '15 at 4:50

Yuval FilmusYuval Filmus

48.7k472145

$endgroup$

add a comment | 

0

0

0

$begingroup$

Hint: when $x neq 14$ we have
$$
frac{x^2-14x}{x^2-196} = frac{x(x-14)}{(x+14)(x-14)} = frac{x}{x+14}.
$$

share|cite|improve this answer

answered Apr 10 '15 at 4:50

Yuval FilmusYuval Filmus

48.7k472145

$endgroup$

Hint: when $x neq 14$ we have
$$
frac{x^2-14x}{x^2-196} = frac{x(x-14)}{(x+14)(x-14)} = frac{x}{x+14}.
$$

share|cite|improve this answer

answered Apr 10 '15 at 4:50

Yuval FilmusYuval Filmus

48.7k472145

share|cite|improve this answer

share|cite|improve this answer

answered Apr 10 '15 at 4:50

Yuval FilmusYuval Filmus

48.7k472145

answered Apr 10 '15 at 4:50

Yuval FilmusYuval Filmus

48.7k472145

answered Apr 10 '15 at 4:50

Yuval FilmusYuval Filmus

48.7k472145

48.7k472145

add a comment | 

add a comment | 

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Use MathJax to format equations. MathJax reference.

To learn more, see our tips on writing great answers.

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