Finding $lim_{xtopi/2}left(frac{1-sin x}{(pi-2x)^4}right)(cos x)(8x^3 - pi^3)$ using algebra of limits
Let $$
F(x) = left(frac{1-sin x}{(pi-2x)^4}right)(cos x)(8x^3 - pi^3)
$$
Then find the limit of $F(x)$ as $x$ tends to $pi/2$.
How can we find the limit using algebra of limits?
The limit of $dfrac{1-sin x}{(pi-2x)^4}$ as $x$ tends to $pi/2$ is some non-zero finite number. And the limit of rest part is zero. Then why the limit of the whole function is not $0$?
The limit of $F(x)$ is $-frac{3pi^2}{16}$.
How to apply algebra of limits here?
calculus
edited Nov 26 '18 at 8:26
Blue
47.6k870151
asked Nov 26 '18 at 7:37
Mathsaddict
2458
add a comment |
Let $$
F(x) = left(frac{1-sin x}{(pi-2x)^4}right)(cos x)(8x^3 - pi^3)
$$
Then find the limit of $F(x)$ as $x$ tends to $pi/2$.
How can we find the limit using algebra of limits?
The limit of $dfrac{1-sin x}{(pi-2x)^4}$ as $x$ tends to $pi/2$ is some non-zero finite number. And the limit of rest part is zero. Then why the limit of the whole function is not $0$?
The limit of $F(x)$ is $-frac{3pi^2}{16}$.
How to apply algebra of limits here?
calculus
edited Nov 26 '18 at 8:26
Blue
47.6k870151
asked Nov 26 '18 at 7:37
Mathsaddict
2458
add a comment |
Let $$
F(x) = left(frac{1-sin x}{(pi-2x)^4}right)(cos x)(8x^3 - pi^3)
$$
Then find the limit of $F(x)$ as $x$ tends to $pi/2$.
How can we find the limit using algebra of limits?
The limit of $dfrac{1-sin x}{(pi-2x)^4}$ as $x$ tends to $pi/2$ is some non-zero finite number. And the limit of rest part is zero. Then why the limit of the whole function is not $0$?
The limit of $F(x)$ is $-frac{3pi^2}{16}$.
How to apply algebra of limits here?
calculus
edited Nov 26 '18 at 8:26
Blue
47.6k870151
asked Nov 26 '18 at 7:37
Mathsaddict
2458
Let $$
F(x) = left(frac{1-sin x}{(pi-2x)^4}right)(cos x)(8x^3 - pi^3)
$$
Then find the limit of $F(x)$ as $x$ tends to $pi/2$.
How can we find the limit using algebra of limits?
The limit of $dfrac{1-sin x}{(pi-2x)^4}$ as $x$ tends to $pi/2$ is some non-zero finite number. And the limit of rest part is zero. Then why the limit of the whole function is not $0$?
The limit of $F(x)$ is $-frac{3pi^2}{16}$.
How to apply algebra of limits here?
calculus
calculus
edited Nov 26 '18 at 8:26
Blue
47.6k870151
asked Nov 26 '18 at 7:37
Mathsaddict
2458
edited Nov 26 '18 at 8:26
Blue
47.6k870151
asked Nov 26 '18 at 7:37
Mathsaddict
2458
edited Nov 26 '18 at 8:26
Blue
47.6k870151
edited Nov 26 '18 at 8:26
Blue
47.6k870151
edited Nov 26 '18 at 8:26
Blue
47.6k870151
47.6k870151
asked Nov 26 '18 at 7:37
Mathsaddict
2458
asked Nov 26 '18 at 7:37
Mathsaddict
2458
asked Nov 26 '18 at 7:37
Mathsaddict
2458
2458
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
The limit of $frac{1-sin x}{(pi-2x)^4}$ as $x$ tends to $pi/2$ is NOT some non zero finite number.
Note that $(8x^3 - pi^3)=(2x-pi)(4x^2+2xpi+pi^2)$, then
$$F(x)=frac{1-sin x}{(2x-pi)^2}cdotfrac{cos x}{(2x-pi)}cdot (4x^2+2xpi+pi^2)$$
Now evaluate the limit of each factor. This time they are all finite numbers and we can apply the rule related to your previous question How to apply algebra of limits?
Can you take it from here?
answered Nov 26 '18 at 8:03
Robert Z
93.4k1061132
add a comment |
Make life simpler using $x=y+frac pi 2$ which makes
$$lim_{xtofrac pi2}left(frac{1-sin (x)}{(pi-2x)^4}right)cos (x)(8x^3 - pi^3)=lim_{yto 0}frac{left(4 y^2+6 pi y+3 pi ^2right) sin (y) (cos (y)-1)}{8 y^3}$$ Now, use the usual Taylor the series and you will get
$$frac{left(4 y^2+6 pi y+3 pi ^2right) sin (y) (cos (y)-1)}{8 y^3}=-frac{3 pi ^2}{16}-frac{3 pi y}{8}+Oleft(y^2right)$$ which shows the limit and how it is approached.
answered Nov 26 '18 at 8:42
Claude Leibovici
119k1157132
add a comment |
Because we are dealing with an indeterminate form $0cdot infty$ and algebraic theorems do not apply.
To solve the limit let $y=pi/2-xto 0$ then
$$frac{1-sin x}{(pi-2x)^4}cos x(8x^3 - pi^3)=-frac{1-cos y}{16y^4}8ysin y left((pi/2-y)^2+(pi/2-y)pi/2+(pi/2)^2right)=$$
$$=-frac{1-cos y}{2y^2}frac{sin y}y left((pi/2-y)^2+(pi/2-y)pi/2+(pi/2)^2right)$$
answered Nov 26 '18 at 7:41
gimusi
1
I didn't get this indeterminate form. Can you please explain.
– Mathsaddict
Nov 26 '18 at 7:47
What about the limit of $frac{(1-sin x}{(pi-2x)^4}$?
– gimusi
Nov 26 '18 at 7:54
Oh, it is infinity. I made mistake in finding this one. Thanks.
– Mathsaddict
Nov 26 '18 at 8:16
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3013994%2ffinding-lim-x-to-pi-2-left-frac1-sin-x-pi-2x4-right-cos-x8x3%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
The limit of $frac{1-sin x}{(pi-2x)^4}$ as $x$ tends to $pi/2$ is NOT some non zero finite number.
Note that $(8x^3 - pi^3)=(2x-pi)(4x^2+2xpi+pi^2)$, then
$$F(x)=frac{1-sin x}{(2x-pi)^2}cdotfrac{cos x}{(2x-pi)}cdot (4x^2+2xpi+pi^2)$$
Now evaluate the limit of each factor. This time they are all finite numbers and we can apply the rule related to your previous question How to apply algebra of limits?
Can you take it from here?
answered Nov 26 '18 at 8:03
Robert Z
93.4k1061132
add a comment |
The limit of $frac{1-sin x}{(pi-2x)^4}$ as $x$ tends to $pi/2$ is NOT some non zero finite number.
Note that $(8x^3 - pi^3)=(2x-pi)(4x^2+2xpi+pi^2)$, then
$$F(x)=frac{1-sin x}{(2x-pi)^2}cdotfrac{cos x}{(2x-pi)}cdot (4x^2+2xpi+pi^2)$$
Now evaluate the limit of each factor. This time they are all finite numbers and we can apply the rule related to your previous question How to apply algebra of limits?
Can you take it from here?
answered Nov 26 '18 at 8:03
Robert Z
93.4k1061132
add a comment |
The limit of $frac{1-sin x}{(pi-2x)^4}$ as $x$ tends to $pi/2$ is NOT some non zero finite number.
Note that $(8x^3 - pi^3)=(2x-pi)(4x^2+2xpi+pi^2)$, then
$$F(x)=frac{1-sin x}{(2x-pi)^2}cdotfrac{cos x}{(2x-pi)}cdot (4x^2+2xpi+pi^2)$$
Now evaluate the limit of each factor. This time they are all finite numbers and we can apply the rule related to your previous question How to apply algebra of limits?
Can you take it from here?
answered Nov 26 '18 at 8:03
Robert Z
93.4k1061132
The limit of $frac{1-sin x}{(pi-2x)^4}$ as $x$ tends to $pi/2$ is NOT some non zero finite number.
Note that $(8x^3 - pi^3)=(2x-pi)(4x^2+2xpi+pi^2)$, then
$$F(x)=frac{1-sin x}{(2x-pi)^2}cdotfrac{cos x}{(2x-pi)}cdot (4x^2+2xpi+pi^2)$$
Now evaluate the limit of each factor. This time they are all finite numbers and we can apply the rule related to your previous question How to apply algebra of limits?
Can you take it from here?
answered Nov 26 '18 at 8:03
Robert Z
93.4k1061132
edited Nov 26 '18 at 8:19
answered Nov 26 '18 at 8:03
Robert Z
93.4k1061132
answered Nov 26 '18 at 8:03
Robert Z
93.4k1061132
answered Nov 26 '18 at 8:03
Robert Z
93.4k1061132
93.4k1061132
add a comment |
add a comment |
Make life simpler using $x=y+frac pi 2$ which makes
$$lim_{xtofrac pi2}left(frac{1-sin (x)}{(pi-2x)^4}right)cos (x)(8x^3 - pi^3)=lim_{yto 0}frac{left(4 y^2+6 pi y+3 pi ^2right) sin (y) (cos (y)-1)}{8 y^3}$$ Now, use the usual Taylor the series and you will get
$$frac{left(4 y^2+6 pi y+3 pi ^2right) sin (y) (cos (y)-1)}{8 y^3}=-frac{3 pi ^2}{16}-frac{3 pi y}{8}+Oleft(y^2right)$$ which shows the limit and how it is approached.
answered Nov 26 '18 at 8:42
Claude Leibovici
119k1157132
add a comment |
Make life simpler using $x=y+frac pi 2$ which makes
$$lim_{xtofrac pi2}left(frac{1-sin (x)}{(pi-2x)^4}right)cos (x)(8x^3 - pi^3)=lim_{yto 0}frac{left(4 y^2+6 pi y+3 pi ^2right) sin (y) (cos (y)-1)}{8 y^3}$$ Now, use the usual Taylor the series and you will get
$$frac{left(4 y^2+6 pi y+3 pi ^2right) sin (y) (cos (y)-1)}{8 y^3}=-frac{3 pi ^2}{16}-frac{3 pi y}{8}+Oleft(y^2right)$$ which shows the limit and how it is approached.
answered Nov 26 '18 at 8:42
Claude Leibovici
119k1157132
add a comment |
Make life simpler using $x=y+frac pi 2$ which makes
$$lim_{xtofrac pi2}left(frac{1-sin (x)}{(pi-2x)^4}right)cos (x)(8x^3 - pi^3)=lim_{yto 0}frac{left(4 y^2+6 pi y+3 pi ^2right) sin (y) (cos (y)-1)}{8 y^3}$$ Now, use the usual Taylor the series and you will get
$$frac{left(4 y^2+6 pi y+3 pi ^2right) sin (y) (cos (y)-1)}{8 y^3}=-frac{3 pi ^2}{16}-frac{3 pi y}{8}+Oleft(y^2right)$$ which shows the limit and how it is approached.
answered Nov 26 '18 at 8:42
Claude Leibovici
119k1157132
Make life simpler using $x=y+frac pi 2$ which makes
$$lim_{xtofrac pi2}left(frac{1-sin (x)}{(pi-2x)^4}right)cos (x)(8x^3 - pi^3)=lim_{yto 0}frac{left(4 y^2+6 pi y+3 pi ^2right) sin (y) (cos (y)-1)}{8 y^3}$$ Now, use the usual Taylor the series and you will get
$$frac{left(4 y^2+6 pi y+3 pi ^2right) sin (y) (cos (y)-1)}{8 y^3}=-frac{3 pi ^2}{16}-frac{3 pi y}{8}+Oleft(y^2right)$$ which shows the limit and how it is approached.
answered Nov 26 '18 at 8:42
Claude Leibovici
119k1157132
answered Nov 26 '18 at 8:42
Claude Leibovici
119k1157132
answered Nov 26 '18 at 8:42
Claude Leibovici
119k1157132
answered Nov 26 '18 at 8:42
Claude Leibovici
119k1157132
119k1157132
add a comment |
add a comment |
Because we are dealing with an indeterminate form $0cdot infty$ and algebraic theorems do not apply.
To solve the limit let $y=pi/2-xto 0$ then
$$frac{1-sin x}{(pi-2x)^4}cos x(8x^3 - pi^3)=-frac{1-cos y}{16y^4}8ysin y left((pi/2-y)^2+(pi/2-y)pi/2+(pi/2)^2right)=$$
$$=-frac{1-cos y}{2y^2}frac{sin y}y left((pi/2-y)^2+(pi/2-y)pi/2+(pi/2)^2right)$$
answered Nov 26 '18 at 7:41
gimusi
1
I didn't get this indeterminate form. Can you please explain.
– Mathsaddict
Nov 26 '18 at 7:47
What about the limit of $frac{(1-sin x}{(pi-2x)^4}$?
– gimusi
Nov 26 '18 at 7:54
Oh, it is infinity. I made mistake in finding this one. Thanks.
– Mathsaddict
Nov 26 '18 at 8:16
add a comment |
Because we are dealing with an indeterminate form $0cdot infty$ and algebraic theorems do not apply.
To solve the limit let $y=pi/2-xto 0$ then
$$frac{1-sin x}{(pi-2x)^4}cos x(8x^3 - pi^3)=-frac{1-cos y}{16y^4}8ysin y left((pi/2-y)^2+(pi/2-y)pi/2+(pi/2)^2right)=$$
$$=-frac{1-cos y}{2y^2}frac{sin y}y left((pi/2-y)^2+(pi/2-y)pi/2+(pi/2)^2right)$$
answered Nov 26 '18 at 7:41
gimusi
1
I didn't get this indeterminate form. Can you please explain.
– Mathsaddict
Nov 26 '18 at 7:47
What about the limit of $frac{(1-sin x}{(pi-2x)^4}$?
– gimusi
Nov 26 '18 at 7:54
Oh, it is infinity. I made mistake in finding this one. Thanks.
– Mathsaddict
Nov 26 '18 at 8:16
add a comment |
Because we are dealing with an indeterminate form $0cdot infty$ and algebraic theorems do not apply.
To solve the limit let $y=pi/2-xto 0$ then
$$frac{1-sin x}{(pi-2x)^4}cos x(8x^3 - pi^3)=-frac{1-cos y}{16y^4}8ysin y left((pi/2-y)^2+(pi/2-y)pi/2+(pi/2)^2right)=$$
$$=-frac{1-cos y}{2y^2}frac{sin y}y left((pi/2-y)^2+(pi/2-y)pi/2+(pi/2)^2right)$$
answered Nov 26 '18 at 7:41
gimusi
1
Because we are dealing with an indeterminate form $0cdot infty$ and algebraic theorems do not apply.
To solve the limit let $y=pi/2-xto 0$ then
$$frac{1-sin x}{(pi-2x)^4}cos x(8x^3 - pi^3)=-frac{1-cos y}{16y^4}8ysin y left((pi/2-y)^2+(pi/2-y)pi/2+(pi/2)^2right)=$$
$$=-frac{1-cos y}{2y^2}frac{sin y}y left((pi/2-y)^2+(pi/2-y)pi/2+(pi/2)^2right)$$
answered Nov 26 '18 at 7:41
gimusi
1
edited Nov 26 '18 at 7:59
answered Nov 26 '18 at 7:41
gimusi
1
answered Nov 26 '18 at 7:41
gimusi
1
answered Nov 26 '18 at 7:41
gimusi
1
1
I didn't get this indeterminate form. Can you please explain.
– Mathsaddict
Nov 26 '18 at 7:47
What about the limit of $frac{(1-sin x}{(pi-2x)^4}$?
– gimusi
Nov 26 '18 at 7:54
Oh, it is infinity. I made mistake in finding this one. Thanks.
– Mathsaddict
Nov 26 '18 at 8:16
add a comment |
I didn't get this indeterminate form. Can you please explain.
– Mathsaddict
Nov 26 '18 at 7:47
What about the limit of $frac{(1-sin x}{(pi-2x)^4}$?
– gimusi
Nov 26 '18 at 7:54
Oh, it is infinity. I made mistake in finding this one. Thanks.
– Mathsaddict
Nov 26 '18 at 8:16
I didn't get this indeterminate form. Can you please explain.
– Mathsaddict
Nov 26 '18 at 7:47
I didn't get this indeterminate form. Can you please explain.
– Mathsaddict
Nov 26 '18 at 7:47
What about the limit of $frac{(1-sin x}{(pi-2x)^4}$?
– gimusi
Nov 26 '18 at 7:54
What about the limit of $frac{(1-sin x}{(pi-2x)^4}$?
– gimusi
Nov 26 '18 at 7:54
Oh, it is infinity. I made mistake in finding this one. Thanks.
– Mathsaddict
Nov 26 '18 at 8:16
Oh, it is infinity. I made mistake in finding this one. Thanks.
– Mathsaddict
Nov 26 '18 at 8:16
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- 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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3013994%2ffinding-lim-x-to-pi-2-left-frac1-sin-x-pi-2x4-right-cos-x8x3%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
