Evaluating $lim_{x to 0} frac{a_1exp(-b_1x^2)}{sum_i a_iexp(-b_ix^2)}$

0

$begingroup$

I am looking for the solution of the following limit:

$$lim_{x to 0} frac{a_1exp(-b_1x^2)}{sum_i a_iexp(-b_ix^2)}$$

Since $lim_{yto0}exp(y) = 1$, is $frac{a_1}{sum_i a_i}$ really the correct solution for this problem?

limits proof-verification

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edited Jan 21 at 20:26

Blue

48.8k870156

asked Jan 21 at 15:26

Ben LBen L

114

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes that is correct.
  $endgroup$
  – lightxbulb
  Jan 21 at 15:32
  

  
  
  
  

add a comment | 

0

$begingroup$

I am looking for the solution of the following limit:

$$lim_{x to 0} frac{a_1exp(-b_1x^2)}{sum_i a_iexp(-b_ix^2)}$$

Since $lim_{yto0}exp(y) = 1$, is $frac{a_1}{sum_i a_i}$ really the correct solution for this problem?

limits proof-verification

share|cite|improve this question

edited Jan 21 at 20:26

Blue

48.8k870156

asked Jan 21 at 15:26

Ben LBen L

114

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes that is correct.
  $endgroup$
  – lightxbulb
  Jan 21 at 15:32
  

  
  
  
  

add a comment | 

0

0

0

$begingroup$

I am looking for the solution of the following limit:

$$lim_{x to 0} frac{a_1exp(-b_1x^2)}{sum_i a_iexp(-b_ix^2)}$$

Since $lim_{yto0}exp(y) = 1$, is $frac{a_1}{sum_i a_i}$ really the correct solution for this problem?

limits proof-verification

share|cite|improve this question

edited Jan 21 at 20:26

Blue

48.8k870156

asked Jan 21 at 15:26

Ben LBen L

114

$endgroup$

I am looking for the solution of the following limit:

$$lim_{x to 0} frac{a_1exp(-b_1x^2)}{sum_i a_iexp(-b_ix^2)}$$

Since $lim_{yto0}exp(y) = 1$, is $frac{a_1}{sum_i a_i}$ really the correct solution for this problem?

limits proof-verification

limits proof-verification

share|cite|improve this question

edited Jan 21 at 20:26

Blue

48.8k870156

asked Jan 21 at 15:26

Ben LBen L

114

share|cite|improve this question

edited Jan 21 at 20:26

Blue

48.8k870156

asked Jan 21 at 15:26

Ben LBen L

114

share|cite|improve this question

share|cite|improve this question

edited Jan 21 at 20:26

Blue

48.8k870156

edited Jan 21 at 20:26

Blue

48.8k870156

edited Jan 21 at 20:26

Blue

48.8k870156

48.8k870156

asked Jan 21 at 15:26

Ben LBen L

114

asked Jan 21 at 15:26

Ben LBen L

114

asked Jan 21 at 15:26

Ben LBen L

114

114

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes that is correct.
  $endgroup$
  – lightxbulb
  Jan 21 at 15:32
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes that is correct.
  $endgroup$
  – lightxbulb
  Jan 21 at 15:32
  

  
  
  
  

$begingroup$
Yes that is correct.
$endgroup$
– lightxbulb
Jan 21 at 15:32

$begingroup$
Yes that is correct.
$endgroup$
– lightxbulb
Jan 21 at 15:32

add a comment | 

1 Answer
1

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1

$begingroup$

Provided the limit exists for $f,g$, the latter being nonzero, we can say

$$lim limits_{x to c} frac{f(x)}{g(x)} = frac{lim limits_{x to c} f(x)}{lim limits_{x to c} g(x)}$$

Take

• $c = 0$


• $f(x) = a_1exp(-b_1x^2)$


• 
  $g(x) = sum limits_{i} a_iexp(-b_ix^2) = exp(-b_ix^2) sum limits_{i} a_i$.

Thus, clearly,

$$begin{align}
lim limits_{x to c} f(x) &= a_1 \
lim limits_{x to c} g(x) &= sum limits_{i} a_i
end{align}$$

ultimately verifying your result, if the limit of the denominator can be guaranteed to be nonzero.

share|cite|improve this answer

edited Jan 21 at 20:23

answered Jan 21 at 15:37

Eevee TrainerEevee Trainer

7,51721338

$endgroup$

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

1

$begingroup$

Provided the limit exists for $f,g$, the latter being nonzero, we can say

$$lim limits_{x to c} frac{f(x)}{g(x)} = frac{lim limits_{x to c} f(x)}{lim limits_{x to c} g(x)}$$

Take

• $c = 0$


• $f(x) = a_1exp(-b_1x^2)$


• 
  $g(x) = sum limits_{i} a_iexp(-b_ix^2) = exp(-b_ix^2) sum limits_{i} a_i$.

Thus, clearly,

$$begin{align}
lim limits_{x to c} f(x) &= a_1 \
lim limits_{x to c} g(x) &= sum limits_{i} a_i
end{align}$$

ultimately verifying your result, if the limit of the denominator can be guaranteed to be nonzero.

share|cite|improve this answer

edited Jan 21 at 20:23

answered Jan 21 at 15:37

Eevee TrainerEevee Trainer

7,51721338

$endgroup$

add a comment | 

1

$begingroup$

Provided the limit exists for $f,g$, the latter being nonzero, we can say

$$lim limits_{x to c} frac{f(x)}{g(x)} = frac{lim limits_{x to c} f(x)}{lim limits_{x to c} g(x)}$$

Take

• $c = 0$


• $f(x) = a_1exp(-b_1x^2)$


• 
  $g(x) = sum limits_{i} a_iexp(-b_ix^2) = exp(-b_ix^2) sum limits_{i} a_i$.

Thus, clearly,

$$begin{align}
lim limits_{x to c} f(x) &= a_1 \
lim limits_{x to c} g(x) &= sum limits_{i} a_i
end{align}$$

ultimately verifying your result, if the limit of the denominator can be guaranteed to be nonzero.

share|cite|improve this answer

edited Jan 21 at 20:23

answered Jan 21 at 15:37

Eevee TrainerEevee Trainer

7,51721338

$endgroup$

add a comment | 

1

1

1

$begingroup$

Provided the limit exists for $f,g$, the latter being nonzero, we can say

$$lim limits_{x to c} frac{f(x)}{g(x)} = frac{lim limits_{x to c} f(x)}{lim limits_{x to c} g(x)}$$

Take

• $c = 0$


• $f(x) = a_1exp(-b_1x^2)$


• 
  $g(x) = sum limits_{i} a_iexp(-b_ix^2) = exp(-b_ix^2) sum limits_{i} a_i$.

Thus, clearly,

$$begin{align}
lim limits_{x to c} f(x) &= a_1 \
lim limits_{x to c} g(x) &= sum limits_{i} a_i
end{align}$$

ultimately verifying your result, if the limit of the denominator can be guaranteed to be nonzero.

share|cite|improve this answer

edited Jan 21 at 20:23

answered Jan 21 at 15:37

Eevee TrainerEevee Trainer

7,51721338

$endgroup$

Provided the limit exists for $f,g$, the latter being nonzero, we can say

$$lim limits_{x to c} frac{f(x)}{g(x)} = frac{lim limits_{x to c} f(x)}{lim limits_{x to c} g(x)}$$

Take

• $c = 0$


• $f(x) = a_1exp(-b_1x^2)$


• 
  $g(x) = sum limits_{i} a_iexp(-b_ix^2) = exp(-b_ix^2) sum limits_{i} a_i$.

Thus, clearly,

$$begin{align}
lim limits_{x to c} f(x) &= a_1 \
lim limits_{x to c} g(x) &= sum limits_{i} a_i
end{align}$$

ultimately verifying your result, if the limit of the denominator can be guaranteed to be nonzero.

share|cite|improve this answer

edited Jan 21 at 20:23

answered Jan 21 at 15:37

Eevee TrainerEevee Trainer

7,51721338

share|cite|improve this answer

share|cite|improve this answer

edited Jan 21 at 20:23

edited Jan 21 at 20:23

edited Jan 21 at 20:23

answered Jan 21 at 15:37

Eevee TrainerEevee Trainer

7,51721338

answered Jan 21 at 15:37

Eevee TrainerEevee Trainer

7,51721338

answered Jan 21 at 15:37

Eevee TrainerEevee Trainer

7,51721338

7,51721338

add a comment | 

add a comment | 

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• Asking for help, clarification, or responding to other answers.


• Making statements based on opinion; back them up with references or personal experience.

Use MathJax to format equations. MathJax reference.

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

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• Making statements based on opinion; back them up with references or personal experience.

Use MathJax to format equations. MathJax reference.

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