What is $lim_{x to 3} (3^{x-2}-3)/(x-3)(x+5)$ without l'Hôpital's rule?
up vote
1
down vote
favorite
I'm trying to solve the limit $lim_{x to 3} frac{3^{x-2}-3}{(x-3)(x+5)}$
but I don't know how to proceed: $lim_{x to 3} frac{1}{x+5}$ $lim_{x to 3} frac{3^{x-2}-3}{x-3}$ = $1over8$ $lim_{x to 3} frac{frac{1}{9}(3^{x}-27)}{x-3}$
Any hints? Thanks in advance.
calculus limits functions exponential-function limits-without-lhopital
asked Nov 21 at 15:59
Aleph_0
325
|
show 1 more comment
up vote
1
down vote
favorite
I'm trying to solve the limit $lim_{x to 3} frac{3^{x-2}-3}{(x-3)(x+5)}$
but I don't know how to proceed: $lim_{x to 3} frac{1}{x+5}$ $lim_{x to 3} frac{3^{x-2}-3}{x-3}$ = $1over8$ $lim_{x to 3} frac{frac{1}{9}(3^{x}-27)}{x-3}$
Any hints? Thanks in advance.
calculus limits functions exponential-function limits-without-lhopital
asked Nov 21 at 15:59
Aleph_0
325
How does $3^{x-2}-3$ become $frac1{27}(3^x-1)$?
– Henning Makholm
Nov 21 at 16:07
Note that $3^{x-2}-3ne (3^x-1)/27$
– Andrei
Nov 21 at 16:07
Also, when you say "without l'Hospital", do you mean to exclude any approach that will require you to differentiate the exponential symbolically, or just the literal mention of l'Hospital?
– Henning Makholm
Nov 21 at 16:08
I'm sorry, I intended to write $frac{frac{1}{9}(3^{x}-27)}{x-3}$
– Aleph_0
Nov 21 at 16:17
Yeah, I mean to exclude any approach that requires to differentiate, hence using l'Hospital's theorem
– Aleph_0
Nov 21 at 16:18
|
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm trying to solve the limit $lim_{x to 3} frac{3^{x-2}-3}{(x-3)(x+5)}$
but I don't know how to proceed: $lim_{x to 3} frac{1}{x+5}$ $lim_{x to 3} frac{3^{x-2}-3}{x-3}$ = $1over8$ $lim_{x to 3} frac{frac{1}{9}(3^{x}-27)}{x-3}$
Any hints? Thanks in advance.
calculus limits functions exponential-function limits-without-lhopital
asked Nov 21 at 15:59
Aleph_0
325
I'm trying to solve the limit $lim_{x to 3} frac{3^{x-2}-3}{(x-3)(x+5)}$
but I don't know how to proceed: $lim_{x to 3} frac{1}{x+5}$ $lim_{x to 3} frac{3^{x-2}-3}{x-3}$ = $1over8$ $lim_{x to 3} frac{frac{1}{9}(3^{x}-27)}{x-3}$
Any hints? Thanks in advance.
calculus limits functions exponential-function limits-without-lhopital
calculus limits functions exponential-function limits-without-lhopital
asked Nov 21 at 15:59
Aleph_0
325
asked Nov 21 at 15:59
Aleph_0
325
edited Nov 21 at 16:17
asked Nov 21 at 15:59
Aleph_0
325
asked Nov 21 at 15:59
Aleph_0
325
asked Nov 21 at 15:59
Aleph_0
325
325
How does $3^{x-2}-3$ become $frac1{27}(3^x-1)$?
– Henning Makholm
Nov 21 at 16:07
Note that $3^{x-2}-3ne (3^x-1)/27$
– Andrei
Nov 21 at 16:07
Also, when you say "without l'Hospital", do you mean to exclude any approach that will require you to differentiate the exponential symbolically, or just the literal mention of l'Hospital?
– Henning Makholm
Nov 21 at 16:08
I'm sorry, I intended to write $frac{frac{1}{9}(3^{x}-27)}{x-3}$
– Aleph_0
Nov 21 at 16:17
Yeah, I mean to exclude any approach that requires to differentiate, hence using l'Hospital's theorem
– Aleph_0
Nov 21 at 16:18
|
show 1 more comment
How does $3^{x-2}-3$ become $frac1{27}(3^x-1)$?
– Henning Makholm
Nov 21 at 16:07
Note that $3^{x-2}-3ne (3^x-1)/27$
– Andrei
Nov 21 at 16:07
Also, when you say "without l'Hospital", do you mean to exclude any approach that will require you to differentiate the exponential symbolically, or just the literal mention of l'Hospital?
– Henning Makholm
Nov 21 at 16:08
I'm sorry, I intended to write $frac{frac{1}{9}(3^{x}-27)}{x-3}$
– Aleph_0
Nov 21 at 16:17
Yeah, I mean to exclude any approach that requires to differentiate, hence using l'Hospital's theorem
– Aleph_0
Nov 21 at 16:18
How does $3^{x-2}-3$ become $frac1{27}(3^x-1)$?
– Henning Makholm
Nov 21 at 16:07
How does $3^{x-2}-3$ become $frac1{27}(3^x-1)$?
– Henning Makholm
Nov 21 at 16:07
Note that $3^{x-2}-3ne (3^x-1)/27$
– Andrei
Nov 21 at 16:07
Note that $3^{x-2}-3ne (3^x-1)/27$
– Andrei
Nov 21 at 16:07
Also, when you say "without l'Hospital", do you mean to exclude any approach that will require you to differentiate the exponential symbolically, or just the literal mention of l'Hospital?
– Henning Makholm
Nov 21 at 16:08
Also, when you say "without l'Hospital", do you mean to exclude any approach that will require you to differentiate the exponential symbolically, or just the literal mention of l'Hospital?
– Henning Makholm
Nov 21 at 16:08
I'm sorry, I intended to write $frac{frac{1}{9}(3^{x}-27)}{x-3}$
– Aleph_0
Nov 21 at 16:17
I'm sorry, I intended to write $frac{frac{1}{9}(3^{x}-27)}{x-3}$
– Aleph_0
Nov 21 at 16:17
Yeah, I mean to exclude any approach that requires to differentiate, hence using l'Hospital's theorem
– Aleph_0
Nov 21 at 16:18
Yeah, I mean to exclude any approach that requires to differentiate, hence using l'Hospital's theorem
– Aleph_0
Nov 21 at 16:18
|
show 1 more comment
4 Answers
4
active
oldest
votes
up vote
2
down vote
accepted
We have that with $y=x-3 to 0$
$$lim_{x to 3} frac{3^{x-2}-3}{(x-3)(x+5)}=lim_{y to 0} frac{3cdot 3^{y}-3}{y(y+8)}=lim_{y to 0} frac{3}{y+8}frac{3^{y}-1}{y}$$
then recall that by standard limits
$$frac{3^{y}-1}{y}=frac{e^{ylog 3}-1}{ylog 3}cdot log 3 to log 3$$
answered Nov 21 at 16:07
gimusi
1
add a comment |
up vote
4
down vote
Hint:
Apply the definition of derivative
$$lim_{xto3}dfrac{3^{x-2}-3}{x-3}=3lim_{xto3}dfrac{3^{x-3}-1}{x-3}=3(3^{x-3})'Big|_{x=3}=3ln3$$
answered Nov 21 at 16:07
Nosrati
26.4k62353
add a comment |
up vote
2
down vote
Hint:
Set $h=x-3$; $h$ tends to $0$ as $x$ tends to $3$, and you can rewrite the fraction as
$$frac1{h+8},frac{3^{h+1}-3}h=underbrace{frac 3{h+8}}_{substack{downarrow\tfrac38}}underbrace{frac{3^h-1}h}_{substack{downarrow\(3^x)'_{x=0}}}.$$
answered Nov 21 at 16:13
Bernard
117k637110
add a comment |
up vote
1
down vote
Perhaps use definition of $3^x$ ... namely $3^x = e^{xlog 3}$. Instaed of $x to 3$ write $y=x-3$ and do $y to 0$.
$$
lim_{x to 3} frac{3^{x-2}-3}{(x-3)(x+5)} = lim_{y to 0}frac{3^{y+1}-3}{y(y+8)}
=3 lim_{y to 0}frac{3^{y}-1}{y(y+8)}
\
3^y = exp(ylog 3) = 1 + ylog 3 + o(y)
\
3^y-1 = ylog 3 + o(y)
\
y(y+8) = 8y+o(y)
\
frac{1}{y(y+8)} = y^{-1}frac{1}{8}+o(y^{-1})
\
frac{3^y-1}{y(y+8)} = frac{log 3}{8} + o(y)
\
3frac{3^y-1}{y(y+8)} = frac{3log 3}{8} + o(1)
\
3 lim_{y to 0}frac{3^{y}-1}{y(y+8)} = frac{3log 3}{8}
$$
answered Nov 21 at 16:53
GEdgar
61.5k267168
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%2f3007918%2fwhat-is-lim-x-to-3-3x-2-3-x-3x5-without-lh%25c3%25b4pitals-rule%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
We have that with $y=x-3 to 0$
$$lim_{x to 3} frac{3^{x-2}-3}{(x-3)(x+5)}=lim_{y to 0} frac{3cdot 3^{y}-3}{y(y+8)}=lim_{y to 0} frac{3}{y+8}frac{3^{y}-1}{y}$$
then recall that by standard limits
$$frac{3^{y}-1}{y}=frac{e^{ylog 3}-1}{ylog 3}cdot log 3 to log 3$$
answered Nov 21 at 16:07
gimusi
1
add a comment |
up vote
2
down vote
accepted
We have that with $y=x-3 to 0$
$$lim_{x to 3} frac{3^{x-2}-3}{(x-3)(x+5)}=lim_{y to 0} frac{3cdot 3^{y}-3}{y(y+8)}=lim_{y to 0} frac{3}{y+8}frac{3^{y}-1}{y}$$
then recall that by standard limits
$$frac{3^{y}-1}{y}=frac{e^{ylog 3}-1}{ylog 3}cdot log 3 to log 3$$
answered Nov 21 at 16:07
gimusi
1
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
We have that with $y=x-3 to 0$
$$lim_{x to 3} frac{3^{x-2}-3}{(x-3)(x+5)}=lim_{y to 0} frac{3cdot 3^{y}-3}{y(y+8)}=lim_{y to 0} frac{3}{y+8}frac{3^{y}-1}{y}$$
then recall that by standard limits
$$frac{3^{y}-1}{y}=frac{e^{ylog 3}-1}{ylog 3}cdot log 3 to log 3$$
answered Nov 21 at 16:07
gimusi
1
We have that with $y=x-3 to 0$
$$lim_{x to 3} frac{3^{x-2}-3}{(x-3)(x+5)}=lim_{y to 0} frac{3cdot 3^{y}-3}{y(y+8)}=lim_{y to 0} frac{3}{y+8}frac{3^{y}-1}{y}$$
then recall that by standard limits
$$frac{3^{y}-1}{y}=frac{e^{ylog 3}-1}{ylog 3}cdot log 3 to log 3$$
answered Nov 21 at 16:07
gimusi
1
edited Nov 22 at 9:18
answered Nov 21 at 16:07
gimusi
1
answered Nov 21 at 16:07
gimusi
1
answered Nov 21 at 16:07
gimusi
1
1
add a comment |
add a comment |
up vote
4
down vote
Hint:
Apply the definition of derivative
$$lim_{xto3}dfrac{3^{x-2}-3}{x-3}=3lim_{xto3}dfrac{3^{x-3}-1}{x-3}=3(3^{x-3})'Big|_{x=3}=3ln3$$
answered Nov 21 at 16:07
Nosrati
26.4k62353
add a comment |
up vote
4
down vote
Hint:
Apply the definition of derivative
$$lim_{xto3}dfrac{3^{x-2}-3}{x-3}=3lim_{xto3}dfrac{3^{x-3}-1}{x-3}=3(3^{x-3})'Big|_{x=3}=3ln3$$
answered Nov 21 at 16:07
Nosrati
26.4k62353
add a comment |
up vote
4
down vote
up vote
4
down vote
Hint:
Apply the definition of derivative
$$lim_{xto3}dfrac{3^{x-2}-3}{x-3}=3lim_{xto3}dfrac{3^{x-3}-1}{x-3}=3(3^{x-3})'Big|_{x=3}=3ln3$$
answered Nov 21 at 16:07
Nosrati
26.4k62353
Hint:
Apply the definition of derivative
$$lim_{xto3}dfrac{3^{x-2}-3}{x-3}=3lim_{xto3}dfrac{3^{x-3}-1}{x-3}=3(3^{x-3})'Big|_{x=3}=3ln3$$
answered Nov 21 at 16:07
Nosrati
26.4k62353
edited Nov 21 at 16:16
answered Nov 21 at 16:07
Nosrati
26.4k62353
answered Nov 21 at 16:07
Nosrati
26.4k62353
answered Nov 21 at 16:07
Nosrati
26.4k62353
26.4k62353
add a comment |
add a comment |
up vote
2
down vote
Hint:
Set $h=x-3$; $h$ tends to $0$ as $x$ tends to $3$, and you can rewrite the fraction as
$$frac1{h+8},frac{3^{h+1}-3}h=underbrace{frac 3{h+8}}_{substack{downarrow\tfrac38}}underbrace{frac{3^h-1}h}_{substack{downarrow\(3^x)'_{x=0}}}.$$
answered Nov 21 at 16:13
Bernard
117k637110
add a comment |
up vote
2
down vote
Hint:
Set $h=x-3$; $h$ tends to $0$ as $x$ tends to $3$, and you can rewrite the fraction as
$$frac1{h+8},frac{3^{h+1}-3}h=underbrace{frac 3{h+8}}_{substack{downarrow\tfrac38}}underbrace{frac{3^h-1}h}_{substack{downarrow\(3^x)'_{x=0}}}.$$
answered Nov 21 at 16:13
Bernard
117k637110
add a comment |
up vote
2
down vote
up vote
2
down vote
Hint:
Set $h=x-3$; $h$ tends to $0$ as $x$ tends to $3$, and you can rewrite the fraction as
$$frac1{h+8},frac{3^{h+1}-3}h=underbrace{frac 3{h+8}}_{substack{downarrow\tfrac38}}underbrace{frac{3^h-1}h}_{substack{downarrow\(3^x)'_{x=0}}}.$$
answered Nov 21 at 16:13
Bernard
117k637110
Hint:
Set $h=x-3$; $h$ tends to $0$ as $x$ tends to $3$, and you can rewrite the fraction as
$$frac1{h+8},frac{3^{h+1}-3}h=underbrace{frac 3{h+8}}_{substack{downarrow\tfrac38}}underbrace{frac{3^h-1}h}_{substack{downarrow\(3^x)'_{x=0}}}.$$
answered Nov 21 at 16:13
Bernard
117k637110
answered Nov 21 at 16:13
Bernard
117k637110
answered Nov 21 at 16:13
Bernard
117k637110
answered Nov 21 at 16:13
Bernard
117k637110
117k637110
add a comment |
add a comment |
up vote
1
down vote
Perhaps use definition of $3^x$ ... namely $3^x = e^{xlog 3}$. Instaed of $x to 3$ write $y=x-3$ and do $y to 0$.
$$
lim_{x to 3} frac{3^{x-2}-3}{(x-3)(x+5)} = lim_{y to 0}frac{3^{y+1}-3}{y(y+8)}
=3 lim_{y to 0}frac{3^{y}-1}{y(y+8)}
\
3^y = exp(ylog 3) = 1 + ylog 3 + o(y)
\
3^y-1 = ylog 3 + o(y)
\
y(y+8) = 8y+o(y)
\
frac{1}{y(y+8)} = y^{-1}frac{1}{8}+o(y^{-1})
\
frac{3^y-1}{y(y+8)} = frac{log 3}{8} + o(y)
\
3frac{3^y-1}{y(y+8)} = frac{3log 3}{8} + o(1)
\
3 lim_{y to 0}frac{3^{y}-1}{y(y+8)} = frac{3log 3}{8}
$$
answered Nov 21 at 16:53
GEdgar
61.5k267168
add a comment |
up vote
1
down vote
Perhaps use definition of $3^x$ ... namely $3^x = e^{xlog 3}$. Instaed of $x to 3$ write $y=x-3$ and do $y to 0$.
$$
lim_{x to 3} frac{3^{x-2}-3}{(x-3)(x+5)} = lim_{y to 0}frac{3^{y+1}-3}{y(y+8)}
=3 lim_{y to 0}frac{3^{y}-1}{y(y+8)}
\
3^y = exp(ylog 3) = 1 + ylog 3 + o(y)
\
3^y-1 = ylog 3 + o(y)
\
y(y+8) = 8y+o(y)
\
frac{1}{y(y+8)} = y^{-1}frac{1}{8}+o(y^{-1})
\
frac{3^y-1}{y(y+8)} = frac{log 3}{8} + o(y)
\
3frac{3^y-1}{y(y+8)} = frac{3log 3}{8} + o(1)
\
3 lim_{y to 0}frac{3^{y}-1}{y(y+8)} = frac{3log 3}{8}
$$
answered Nov 21 at 16:53
GEdgar
61.5k267168
add a comment |
up vote
1
down vote
up vote
1
down vote
Perhaps use definition of $3^x$ ... namely $3^x = e^{xlog 3}$. Instaed of $x to 3$ write $y=x-3$ and do $y to 0$.
$$
lim_{x to 3} frac{3^{x-2}-3}{(x-3)(x+5)} = lim_{y to 0}frac{3^{y+1}-3}{y(y+8)}
=3 lim_{y to 0}frac{3^{y}-1}{y(y+8)}
\
3^y = exp(ylog 3) = 1 + ylog 3 + o(y)
\
3^y-1 = ylog 3 + o(y)
\
y(y+8) = 8y+o(y)
\
frac{1}{y(y+8)} = y^{-1}frac{1}{8}+o(y^{-1})
\
frac{3^y-1}{y(y+8)} = frac{log 3}{8} + o(y)
\
3frac{3^y-1}{y(y+8)} = frac{3log 3}{8} + o(1)
\
3 lim_{y to 0}frac{3^{y}-1}{y(y+8)} = frac{3log 3}{8}
$$
answered Nov 21 at 16:53
GEdgar
61.5k267168
Perhaps use definition of $3^x$ ... namely $3^x = e^{xlog 3}$. Instaed of $x to 3$ write $y=x-3$ and do $y to 0$.
$$
lim_{x to 3} frac{3^{x-2}-3}{(x-3)(x+5)} = lim_{y to 0}frac{3^{y+1}-3}{y(y+8)}
=3 lim_{y to 0}frac{3^{y}-1}{y(y+8)}
\
3^y = exp(ylog 3) = 1 + ylog 3 + o(y)
\
3^y-1 = ylog 3 + o(y)
\
y(y+8) = 8y+o(y)
\
frac{1}{y(y+8)} = y^{-1}frac{1}{8}+o(y^{-1})
\
frac{3^y-1}{y(y+8)} = frac{log 3}{8} + o(y)
\
3frac{3^y-1}{y(y+8)} = frac{3log 3}{8} + o(1)
\
3 lim_{y to 0}frac{3^{y}-1}{y(y+8)} = frac{3log 3}{8}
$$
answered Nov 21 at 16:53
GEdgar
61.5k267168
answered Nov 21 at 16:53
GEdgar
61.5k267168
answered Nov 21 at 16:53
GEdgar
61.5k267168
answered Nov 21 at 16:53
GEdgar
61.5k267168
61.5k267168
add a comment |
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%2f3007918%2fwhat-is-lim-x-to-3-3x-2-3-x-3x5-without-lh%25c3%25b4pitals-rule%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
How does $3^{x-2}-3$ become $frac1{27}(3^x-1)$?
– Henning Makholm
Nov 21 at 16:07
Note that $3^{x-2}-3ne (3^x-1)/27$
– Andrei
Nov 21 at 16:07
Also, when you say "without l'Hospital", do you mean to exclude any approach that will require you to differentiate the exponential symbolically, or just the literal mention of l'Hospital?
– Henning Makholm
Nov 21 at 16:08
I'm sorry, I intended to write $frac{frac{1}{9}(3^{x}-27)}{x-3}$
– Aleph_0
Nov 21 at 16:17
Yeah, I mean to exclude any approach that requires to differentiate, hence using l'Hospital's theorem
– Aleph_0
Nov 21 at 16:18