Using the sequential definition of uniform continuity to show $sin(x)$ is uniformly continuous on $mathbb{R}$
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I want to show $sin(x)$ is uniformly continuous on $mathbb{R}$. Let ${a_{n}}$ and ${b_{n}}$ be sequences such that $lim_{ntoinfty}[b_{n} - a_{n}] = 0$. Then, we need to show $lim_{ntoinfty} |sin(b_{n}) - sin(a_{n})| = 0$. But, I cannot prove this equality. Can someone please help me?
real-analysis inequality uniform-convergence uniform-continuity
asked 1 hour ago
joseph
385
add a comment |
up vote
0
down vote
favorite
I want to show $sin(x)$ is uniformly continuous on $mathbb{R}$. Let ${a_{n}}$ and ${b_{n}}$ be sequences such that $lim_{ntoinfty}[b_{n} - a_{n}] = 0$. Then, we need to show $lim_{ntoinfty} |sin(b_{n}) - sin(a_{n})| = 0$. But, I cannot prove this equality. Can someone please help me?
real-analysis inequality uniform-convergence uniform-continuity
asked 1 hour ago
joseph
385
2
Mean Value Theorem.
– Kavi Rama Murthy
58 mins ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I want to show $sin(x)$ is uniformly continuous on $mathbb{R}$. Let ${a_{n}}$ and ${b_{n}}$ be sequences such that $lim_{ntoinfty}[b_{n} - a_{n}] = 0$. Then, we need to show $lim_{ntoinfty} |sin(b_{n}) - sin(a_{n})| = 0$. But, I cannot prove this equality. Can someone please help me?
real-analysis inequality uniform-convergence uniform-continuity
asked 1 hour ago
joseph
385
I want to show $sin(x)$ is uniformly continuous on $mathbb{R}$. Let ${a_{n}}$ and ${b_{n}}$ be sequences such that $lim_{ntoinfty}[b_{n} - a_{n}] = 0$. Then, we need to show $lim_{ntoinfty} |sin(b_{n}) - sin(a_{n})| = 0$. But, I cannot prove this equality. Can someone please help me?
real-analysis inequality uniform-convergence uniform-continuity
real-analysis inequality uniform-convergence uniform-continuity
asked 1 hour ago
joseph
385
asked 1 hour ago
joseph
385
asked 1 hour ago
joseph
385
asked 1 hour ago
joseph
385
asked 1 hour ago
joseph
385
385
2
Mean Value Theorem.
– Kavi Rama Murthy
58 mins ago
add a comment |
2
Mean Value Theorem.
– Kavi Rama Murthy
58 mins ago
2
2
Mean Value Theorem.
– Kavi Rama Murthy
58 mins ago
Mean Value Theorem.
– Kavi Rama Murthy
58 mins ago
add a comment |
1 Answer
1
active
oldest
votes
up vote
3
down vote
By the mean value theorem:
$sin(b_{n}) - sin(a_{n})= cos (t_n)(b_n-a_n)$ with $t_n$ between $a_n$ and $b_n$.
Can you proceed ?
answered 57 mins ago
Fred
41.8k1642
i understand now, thanks.
– joseph
52 mins ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
By the mean value theorem:
$sin(b_{n}) - sin(a_{n})= cos (t_n)(b_n-a_n)$ with $t_n$ between $a_n$ and $b_n$.
Can you proceed ?
answered 57 mins ago
Fred
41.8k1642
i understand now, thanks.
– joseph
52 mins ago
add a comment |
up vote
3
down vote
By the mean value theorem:
$sin(b_{n}) - sin(a_{n})= cos (t_n)(b_n-a_n)$ with $t_n$ between $a_n$ and $b_n$.
Can you proceed ?
answered 57 mins ago
Fred
41.8k1642
i understand now, thanks.
– joseph
52 mins ago
add a comment |
up vote
3
down vote
up vote
3
down vote
By the mean value theorem:
$sin(b_{n}) - sin(a_{n})= cos (t_n)(b_n-a_n)$ with $t_n$ between $a_n$ and $b_n$.
Can you proceed ?
answered 57 mins ago
Fred
41.8k1642
By the mean value theorem:
$sin(b_{n}) - sin(a_{n})= cos (t_n)(b_n-a_n)$ with $t_n$ between $a_n$ and $b_n$.
Can you proceed ?
answered 57 mins ago
Fred
41.8k1642
answered 57 mins ago
Fred
41.8k1642
answered 57 mins ago
Fred
41.8k1642
answered 57 mins ago
Fred
41.8k1642
41.8k1642
i understand now, thanks.
– joseph
52 mins ago
add a comment |
i understand now, thanks.
– joseph
52 mins ago
i understand now, thanks.
– joseph
52 mins ago
i understand now, thanks.
– joseph
52 mins ago
add a comment |
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Mean Value Theorem.
– Kavi Rama Murthy
58 mins ago