Mixing
Evaluating $lim_{x to 0} frac{a_1exp(-b_1x^2)}{sum_i a_iexp(-b_ix^2)}$
$begingroup$
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
edited Jan 21 at 20:26
Blue
48.8k870156
asked Jan 21 at 15:26
Ben LBen L
114
$endgroup$
add a comment |
$begingroup$
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
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 |
$begingroup$
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
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
edited Jan 21 at 20:26
Blue
48.8k870156
asked Jan 21 at 15:26
Ben LBen L
114
edited Jan 21 at 20:26
Blue
48.8k870156
asked Jan 21 at 15:26
Ben LBen L
114
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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votes
$begingroup$
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.
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
$begingroup$
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.
answered Jan 21 at 15:37
Eevee TrainerEevee Trainer
7,51721338
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 21 at 15:37
Eevee TrainerEevee Trainer
7,51721338
$endgroup$
add a comment |
$begingroup$
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.
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.
answered Jan 21 at 15:37
Eevee TrainerEevee Trainer
7,51721338
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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$begingroup$
Yes that is correct.
$endgroup$
– lightxbulb
Jan 21 at 15:32