Does $lim_{ntoinfty}frac{3^n+5^n}{(-2)^n+7^n}$ exist?
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My textbook says that this limit doesn't exist, but I don't understand - why? I tried calculating it by taking from both numerator and denominator factors that diverge to $infty$ the fastest:
$$lim_{ntoinfty}frac{3^n+5^n}{(-2)^n+7^n}= lim_{ntoinfty}frac{5^n}{7^n}=0$$
Did I do something wrong here?
limits
asked Dec 3 '18 at 16:38
agromekagromek
345
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add a comment |
$begingroup$
My textbook says that this limit doesn't exist, but I don't understand - why? I tried calculating it by taking from both numerator and denominator factors that diverge to $infty$ the fastest:
$$lim_{ntoinfty}frac{3^n+5^n}{(-2)^n+7^n}= lim_{ntoinfty}frac{5^n}{7^n}=0$$
Did I do something wrong here?
limits
asked Dec 3 '18 at 16:38
agromekagromek
345
$endgroup$
4
$begingroup$
The limit is indeed zero, so it's likely a typo.
$endgroup$
– T. Bongers
Dec 3 '18 at 16:38
2
$begingroup$
You’re correct.
$endgroup$
– KM101
Dec 3 '18 at 16:39
add a comment |
$begingroup$
My textbook says that this limit doesn't exist, but I don't understand - why? I tried calculating it by taking from both numerator and denominator factors that diverge to $infty$ the fastest:
$$lim_{ntoinfty}frac{3^n+5^n}{(-2)^n+7^n}= lim_{ntoinfty}frac{5^n}{7^n}=0$$
Did I do something wrong here?
limits
asked Dec 3 '18 at 16:38
agromekagromek
345
$endgroup$
My textbook says that this limit doesn't exist, but I don't understand - why? I tried calculating it by taking from both numerator and denominator factors that diverge to $infty$ the fastest:
$$lim_{ntoinfty}frac{3^n+5^n}{(-2)^n+7^n}= lim_{ntoinfty}frac{5^n}{7^n}=0$$
Did I do something wrong here?
limits
limits
asked Dec 3 '18 at 16:38
agromekagromek
345
asked Dec 3 '18 at 16:38
agromekagromek
345
asked Dec 3 '18 at 16:38
agromekagromek
345
asked Dec 3 '18 at 16:38
agromekagromek
345
asked Dec 3 '18 at 16:38
agromekagromek
345
345
4
$begingroup$
The limit is indeed zero, so it's likely a typo.
$endgroup$
– T. Bongers
Dec 3 '18 at 16:38
2
$begingroup$
You’re correct.
$endgroup$
– KM101
Dec 3 '18 at 16:39
add a comment |
4
$begingroup$
The limit is indeed zero, so it's likely a typo.
$endgroup$
– T. Bongers
Dec 3 '18 at 16:38
2
$begingroup$
You’re correct.
$endgroup$
– KM101
Dec 3 '18 at 16:39
4
4
$begingroup$
The limit is indeed zero, so it's likely a typo.
$endgroup$
– T. Bongers
Dec 3 '18 at 16:38
$begingroup$
The limit is indeed zero, so it's likely a typo.
$endgroup$
– T. Bongers
Dec 3 '18 at 16:38
2
2
$begingroup$
You’re correct.
$endgroup$
– KM101
Dec 3 '18 at 16:39
$begingroup$
You’re correct.
$endgroup$
– KM101
Dec 3 '18 at 16:39
add a comment |
1 Answer
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$begingroup$
Your result is correct but the way is not so much clear.
To solve properly we can observe that
$$frac{3^n+5^n}{2^n+7^n}le frac{3^n+5^n}{(-2)^n+7^n}le frac{3^n+5^n}{-(2^n)+7^n}$$
and refer to squeeze theorem or more simply dividing by the by leading term $7^n$
$$frac{3^n+5^n}{(-2)^n+7^n}=frac{(3/7)^n+(5/7)^n}{(-2/7)^n+1}$$
answered Dec 3 '18 at 16:40
gimusigimusi
1
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add a comment |
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$begingroup$
Your result is correct but the way is not so much clear.
To solve properly we can observe that
$$frac{3^n+5^n}{2^n+7^n}le frac{3^n+5^n}{(-2)^n+7^n}le frac{3^n+5^n}{-(2^n)+7^n}$$
and refer to squeeze theorem or more simply dividing by the by leading term $7^n$
$$frac{3^n+5^n}{(-2)^n+7^n}=frac{(3/7)^n+(5/7)^n}{(-2/7)^n+1}$$
answered Dec 3 '18 at 16:40
gimusigimusi
1
$endgroup$
add a comment |
$begingroup$
Your result is correct but the way is not so much clear.
To solve properly we can observe that
$$frac{3^n+5^n}{2^n+7^n}le frac{3^n+5^n}{(-2)^n+7^n}le frac{3^n+5^n}{-(2^n)+7^n}$$
and refer to squeeze theorem or more simply dividing by the by leading term $7^n$
$$frac{3^n+5^n}{(-2)^n+7^n}=frac{(3/7)^n+(5/7)^n}{(-2/7)^n+1}$$
answered Dec 3 '18 at 16:40
gimusigimusi
1
$endgroup$
add a comment |
$begingroup$
Your result is correct but the way is not so much clear.
To solve properly we can observe that
$$frac{3^n+5^n}{2^n+7^n}le frac{3^n+5^n}{(-2)^n+7^n}le frac{3^n+5^n}{-(2^n)+7^n}$$
and refer to squeeze theorem or more simply dividing by the by leading term $7^n$
$$frac{3^n+5^n}{(-2)^n+7^n}=frac{(3/7)^n+(5/7)^n}{(-2/7)^n+1}$$
answered Dec 3 '18 at 16:40
gimusigimusi
1
$endgroup$
Your result is correct but the way is not so much clear.
To solve properly we can observe that
$$frac{3^n+5^n}{2^n+7^n}le frac{3^n+5^n}{(-2)^n+7^n}le frac{3^n+5^n}{-(2^n)+7^n}$$
and refer to squeeze theorem or more simply dividing by the by leading term $7^n$
$$frac{3^n+5^n}{(-2)^n+7^n}=frac{(3/7)^n+(5/7)^n}{(-2/7)^n+1}$$
answered Dec 3 '18 at 16:40
gimusigimusi
1
edited Dec 3 '18 at 16:43
answered Dec 3 '18 at 16:40
gimusigimusi
1
answered Dec 3 '18 at 16:40
gimusigimusi
1
answered Dec 3 '18 at 16:40
gimusigimusi
1
1
add a comment |
add a comment |
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4
$begingroup$
The limit is indeed zero, so it's likely a typo.
$endgroup$
– T. Bongers
Dec 3 '18 at 16:38
2
$begingroup$
You’re correct.
$endgroup$
– KM101
Dec 3 '18 at 16:39