The limit $lim_{xto infty}frac{1+cos^2x}{x^2}$ [closed]
$$lim_{xto infty}frac{1+cos^2x}{x^2}$$
I tried the formula $cos^2x = dfrac{1+cos 2x}2.$
But I keep obtaining $dfrac{infty}{infty}$.
calculus limits
edited Nov 27 '18 at 20:23
amWhy
191k28224439
asked Nov 27 '18 at 19:24
Taw Taw
7
closed as off-topic by amWhy, RRL, Saad, Arnaud D., Vidyanshu Mishra Nov 28 '18 at 11:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, RRL, Saad, Arnaud D., Vidyanshu Mishra
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$$lim_{xto infty}frac{1+cos^2x}{x^2}$$
I tried the formula $cos^2x = dfrac{1+cos 2x}2.$
But I keep obtaining $dfrac{infty}{infty}$.
calculus limits
edited Nov 27 '18 at 20:23
amWhy
191k28224439
asked Nov 27 '18 at 19:24
Taw Taw
7
closed as off-topic by amWhy, RRL, Saad, Arnaud D., Vidyanshu Mishra Nov 28 '18 at 11:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, RRL, Saad, Arnaud D., Vidyanshu Mishra
If this question can be reworded to fit the rules in the help center, please edit the question.
3
How do you get $infty$ in the numerator?
– amWhy
Nov 27 '18 at 19:34
add a comment |
$$lim_{xto infty}frac{1+cos^2x}{x^2}$$
I tried the formula $cos^2x = dfrac{1+cos 2x}2.$
But I keep obtaining $dfrac{infty}{infty}$.
calculus limits
edited Nov 27 '18 at 20:23
amWhy
191k28224439
asked Nov 27 '18 at 19:24
Taw Taw
7
$$lim_{xto infty}frac{1+cos^2x}{x^2}$$
I tried the formula $cos^2x = dfrac{1+cos 2x}2.$
But I keep obtaining $dfrac{infty}{infty}$.
calculus limits
calculus limits
edited Nov 27 '18 at 20:23
amWhy
191k28224439
asked Nov 27 '18 at 19:24
Taw Taw
7
edited Nov 27 '18 at 20:23
amWhy
191k28224439
asked Nov 27 '18 at 19:24
Taw Taw
7
edited Nov 27 '18 at 20:23
amWhy
191k28224439
edited Nov 27 '18 at 20:23
amWhy
191k28224439
edited Nov 27 '18 at 20:23
amWhy
191k28224439
191k28224439
asked Nov 27 '18 at 19:24
Taw Taw
7
asked Nov 27 '18 at 19:24
Taw Taw
7
asked Nov 27 '18 at 19:24
Taw Taw
7
7
closed as off-topic by amWhy, RRL, Saad, Arnaud D., Vidyanshu Mishra Nov 28 '18 at 11:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, RRL, Saad, Arnaud D., Vidyanshu Mishra
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, RRL, Saad, Arnaud D., Vidyanshu Mishra Nov 28 '18 at 11:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, RRL, Saad, Arnaud D., Vidyanshu Mishra
If this question can be reworded to fit the rules in the help center, please edit the question.
3
How do you get $infty$ in the numerator?
– amWhy
Nov 27 '18 at 19:34
add a comment |
3
How do you get $infty$ in the numerator?
– amWhy
Nov 27 '18 at 19:34
3
3
How do you get $infty$ in the numerator?
– amWhy
Nov 27 '18 at 19:34
How do you get $infty$ in the numerator?
– amWhy
Nov 27 '18 at 19:34
add a comment |
3 Answers
3
active
oldest
votes
The limit is zero by the squeeze theorem, since $$0<frac{1+cos^2 x}{x^2}leq frac{2}{x^2}$$
answered Nov 27 '18 at 19:26
user376343
2,8382822
2
The right inequality should be $le$.
– gimusi
Nov 27 '18 at 19:32
1
Yes thanks I just returned to fix it.
– user376343
Nov 27 '18 at 19:35
add a comment |
$cos^2 x$ is bounded from $0$ to $1$. Then $$g(x)=frac{1+0}{x^2}lefrac{1+cos^2x}{x^2}lefrac{1+1}{x^2}=h(x)$$
Both $g(x)$ and $h(x)$ tend to $0$ as $xtoinfty$, then by the squeeze theorem $$lim_{xto infty}frac{1+cos^2x}{x^2}=0$$
answered Nov 27 '18 at 19:37
Lorenzo B.
1,8322520
That’s fine, to simplify for the LHS it suffices take $g(x)=0$.
– gimusi
Nov 27 '18 at 19:45
add a comment |
HINT
Simply note that $0le cos^2 xle 1$ and then
$$0le frac{1+cos^2x}{x^2}le frac2{x^2}$$
now take the limit and refer to squeeze theorem.
answered Nov 27 '18 at 19:26
gimusi
1
Please use proper formatting, instead of copying and pasting an ill-formatted OP.
– amWhy
Nov 27 '18 at 19:27
The left hand inequality is strict.
– amWhy
Nov 27 '18 at 19:57
@amWhy The left hand inequality "can also be" strict, we really do not need that. The right hand side can't be strict but it is another story. Please do not hesitate to ask for any other clarification on that!
– gimusi
Nov 27 '18 at 20:02
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
The limit is zero by the squeeze theorem, since $$0<frac{1+cos^2 x}{x^2}leq frac{2}{x^2}$$
answered Nov 27 '18 at 19:26
user376343
2,8382822
2
The right inequality should be $le$.
– gimusi
Nov 27 '18 at 19:32
1
Yes thanks I just returned to fix it.
– user376343
Nov 27 '18 at 19:35
add a comment |
The limit is zero by the squeeze theorem, since $$0<frac{1+cos^2 x}{x^2}leq frac{2}{x^2}$$
answered Nov 27 '18 at 19:26
user376343
2,8382822
2
The right inequality should be $le$.
– gimusi
Nov 27 '18 at 19:32
1
Yes thanks I just returned to fix it.
– user376343
Nov 27 '18 at 19:35
add a comment |
The limit is zero by the squeeze theorem, since $$0<frac{1+cos^2 x}{x^2}leq frac{2}{x^2}$$
answered Nov 27 '18 at 19:26
user376343
2,8382822
The limit is zero by the squeeze theorem, since $$0<frac{1+cos^2 x}{x^2}leq frac{2}{x^2}$$
answered Nov 27 '18 at 19:26
user376343
2,8382822
edited Nov 27 '18 at 19:37
answered Nov 27 '18 at 19:26
user376343
2,8382822
answered Nov 27 '18 at 19:26
user376343
2,8382822
answered Nov 27 '18 at 19:26
user376343
2,8382822
2,8382822
2
The right inequality should be $le$.
– gimusi
Nov 27 '18 at 19:32
1
Yes thanks I just returned to fix it.
– user376343
Nov 27 '18 at 19:35
add a comment |
2
The right inequality should be $le$.
– gimusi
Nov 27 '18 at 19:32
1
Yes thanks I just returned to fix it.
– user376343
Nov 27 '18 at 19:35
2
2
The right inequality should be $le$.
– gimusi
Nov 27 '18 at 19:32
The right inequality should be $le$.
– gimusi
Nov 27 '18 at 19:32
1
1
Yes thanks I just returned to fix it.
– user376343
Nov 27 '18 at 19:35
Yes thanks I just returned to fix it.
– user376343
Nov 27 '18 at 19:35
add a comment |
$cos^2 x$ is bounded from $0$ to $1$. Then $$g(x)=frac{1+0}{x^2}lefrac{1+cos^2x}{x^2}lefrac{1+1}{x^2}=h(x)$$
Both $g(x)$ and $h(x)$ tend to $0$ as $xtoinfty$, then by the squeeze theorem $$lim_{xto infty}frac{1+cos^2x}{x^2}=0$$
answered Nov 27 '18 at 19:37
Lorenzo B.
1,8322520
That’s fine, to simplify for the LHS it suffices take $g(x)=0$.
– gimusi
Nov 27 '18 at 19:45
add a comment |
$cos^2 x$ is bounded from $0$ to $1$. Then $$g(x)=frac{1+0}{x^2}lefrac{1+cos^2x}{x^2}lefrac{1+1}{x^2}=h(x)$$
Both $g(x)$ and $h(x)$ tend to $0$ as $xtoinfty$, then by the squeeze theorem $$lim_{xto infty}frac{1+cos^2x}{x^2}=0$$
answered Nov 27 '18 at 19:37
Lorenzo B.
1,8322520
That’s fine, to simplify for the LHS it suffices take $g(x)=0$.
– gimusi
Nov 27 '18 at 19:45
add a comment |
$cos^2 x$ is bounded from $0$ to $1$. Then $$g(x)=frac{1+0}{x^2}lefrac{1+cos^2x}{x^2}lefrac{1+1}{x^2}=h(x)$$
Both $g(x)$ and $h(x)$ tend to $0$ as $xtoinfty$, then by the squeeze theorem $$lim_{xto infty}frac{1+cos^2x}{x^2}=0$$
answered Nov 27 '18 at 19:37
Lorenzo B.
1,8322520
$cos^2 x$ is bounded from $0$ to $1$. Then $$g(x)=frac{1+0}{x^2}lefrac{1+cos^2x}{x^2}lefrac{1+1}{x^2}=h(x)$$
Both $g(x)$ and $h(x)$ tend to $0$ as $xtoinfty$, then by the squeeze theorem $$lim_{xto infty}frac{1+cos^2x}{x^2}=0$$
answered Nov 27 '18 at 19:37
Lorenzo B.
1,8322520
answered Nov 27 '18 at 19:37
Lorenzo B.
1,8322520
answered Nov 27 '18 at 19:37
Lorenzo B.
1,8322520
answered Nov 27 '18 at 19:37
Lorenzo B.
1,8322520
1,8322520
That’s fine, to simplify for the LHS it suffices take $g(x)=0$.
– gimusi
Nov 27 '18 at 19:45
add a comment |
That’s fine, to simplify for the LHS it suffices take $g(x)=0$.
– gimusi
Nov 27 '18 at 19:45
That’s fine, to simplify for the LHS it suffices take $g(x)=0$.
– gimusi
Nov 27 '18 at 19:45
That’s fine, to simplify for the LHS it suffices take $g(x)=0$.
– gimusi
Nov 27 '18 at 19:45
add a comment |
HINT
Simply note that $0le cos^2 xle 1$ and then
$$0le frac{1+cos^2x}{x^2}le frac2{x^2}$$
now take the limit and refer to squeeze theorem.
answered Nov 27 '18 at 19:26
gimusi
1
Please use proper formatting, instead of copying and pasting an ill-formatted OP.
– amWhy
Nov 27 '18 at 19:27
The left hand inequality is strict.
– amWhy
Nov 27 '18 at 19:57
@amWhy The left hand inequality "can also be" strict, we really do not need that. The right hand side can't be strict but it is another story. Please do not hesitate to ask for any other clarification on that!
– gimusi
Nov 27 '18 at 20:02
add a comment |
HINT
Simply note that $0le cos^2 xle 1$ and then
$$0le frac{1+cos^2x}{x^2}le frac2{x^2}$$
now take the limit and refer to squeeze theorem.
answered Nov 27 '18 at 19:26
gimusi
1
Please use proper formatting, instead of copying and pasting an ill-formatted OP.
– amWhy
Nov 27 '18 at 19:27
The left hand inequality is strict.
– amWhy
Nov 27 '18 at 19:57
@amWhy The left hand inequality "can also be" strict, we really do not need that. The right hand side can't be strict but it is another story. Please do not hesitate to ask for any other clarification on that!
– gimusi
Nov 27 '18 at 20:02
add a comment |
HINT
Simply note that $0le cos^2 xle 1$ and then
$$0le frac{1+cos^2x}{x^2}le frac2{x^2}$$
now take the limit and refer to squeeze theorem.
answered Nov 27 '18 at 19:26
gimusi
1
HINT
Simply note that $0le cos^2 xle 1$ and then
$$0le frac{1+cos^2x}{x^2}le frac2{x^2}$$
now take the limit and refer to squeeze theorem.
answered Nov 27 '18 at 19:26
gimusi
1
edited Nov 27 '18 at 19:39
answered Nov 27 '18 at 19:26
gimusi
1
answered Nov 27 '18 at 19:26
gimusi
1
answered Nov 27 '18 at 19:26
gimusi
1
1
Please use proper formatting, instead of copying and pasting an ill-formatted OP.
– amWhy
Nov 27 '18 at 19:27
The left hand inequality is strict.
– amWhy
Nov 27 '18 at 19:57
@amWhy The left hand inequality "can also be" strict, we really do not need that. The right hand side can't be strict but it is another story. Please do not hesitate to ask for any other clarification on that!
– gimusi
Nov 27 '18 at 20:02
add a comment |
Please use proper formatting, instead of copying and pasting an ill-formatted OP.
– amWhy
Nov 27 '18 at 19:27
The left hand inequality is strict.
– amWhy
Nov 27 '18 at 19:57
@amWhy The left hand inequality "can also be" strict, we really do not need that. The right hand side can't be strict but it is another story. Please do not hesitate to ask for any other clarification on that!
– gimusi
Nov 27 '18 at 20:02
Please use proper formatting, instead of copying and pasting an ill-formatted OP.
– amWhy
Nov 27 '18 at 19:27
Please use proper formatting, instead of copying and pasting an ill-formatted OP.
– amWhy
Nov 27 '18 at 19:27
The left hand inequality is strict.
– amWhy
Nov 27 '18 at 19:57
The left hand inequality is strict.
– amWhy
Nov 27 '18 at 19:57
@amWhy The left hand inequality "can also be" strict, we really do not need that. The right hand side can't be strict but it is another story. Please do not hesitate to ask for any other clarification on that!
– gimusi
Nov 27 '18 at 20:02
@amWhy The left hand inequality "can also be" strict, we really do not need that. The right hand side can't be strict but it is another story. Please do not hesitate to ask for any other clarification on that!
– gimusi
Nov 27 '18 at 20:02
add a comment |
3
How do you get $infty$ in the numerator?
– amWhy
Nov 27 '18 at 19:34