Evaluate $ lim_{xto 0}|frac{5^x - 5^{-x}}{5^x-1}| $ without using L'Hospital's rule.
$begingroup$
$$
lim_{xto 0}left|frac{5^x - 5^{-x}}{5^x-1}right|
$$
I know the limit is equal to 2. But I am not allowed to use L'Hospital.
How can I evaluate the limit without L'Hospital?
calculus limits limits-without-lhopital
edited Nov 30 '18 at 20:57
Martin Sleziak
44.7k9117272
asked Nov 30 '18 at 12:25
nofnof
32
$endgroup$
add a comment |
$begingroup$
$$
lim_{xto 0}left|frac{5^x - 5^{-x}}{5^x-1}right|
$$
I know the limit is equal to 2. But I am not allowed to use L'Hospital.
How can I evaluate the limit without L'Hospital?
calculus limits limits-without-lhopital
edited Nov 30 '18 at 20:57
Martin Sleziak
44.7k9117272
asked Nov 30 '18 at 12:25
nofnof
32
$endgroup$
add a comment |
$begingroup$
$$
lim_{xto 0}left|frac{5^x - 5^{-x}}{5^x-1}right|
$$
I know the limit is equal to 2. But I am not allowed to use L'Hospital.
How can I evaluate the limit without L'Hospital?
calculus limits limits-without-lhopital
edited Nov 30 '18 at 20:57
Martin Sleziak
44.7k9117272
asked Nov 30 '18 at 12:25
nofnof
32
$endgroup$
$$
lim_{xto 0}left|frac{5^x - 5^{-x}}{5^x-1}right|
$$
I know the limit is equal to 2. But I am not allowed to use L'Hospital.
How can I evaluate the limit without L'Hospital?
calculus limits limits-without-lhopital
calculus limits limits-without-lhopital
edited Nov 30 '18 at 20:57
Martin Sleziak
44.7k9117272
asked Nov 30 '18 at 12:25
nofnof
32
edited Nov 30 '18 at 20:57
Martin Sleziak
44.7k9117272
asked Nov 30 '18 at 12:25
nofnof
32
edited Nov 30 '18 at 20:57
Martin Sleziak
44.7k9117272
edited Nov 30 '18 at 20:57
Martin Sleziak
44.7k9117272
edited Nov 30 '18 at 20:57
Martin Sleziak
44.7k9117272
44.7k9117272
asked Nov 30 '18 at 12:25
nofnof
32
asked Nov 30 '18 at 12:25
nofnof
32
asked Nov 30 '18 at 12:25
nofnof
32
32
add a comment |
add a comment |
4 Answers
4
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oldest
votes
$begingroup$
Hint:
$$lim_{x to 0}biggvertfrac{5^x-5^{-x}}{5^x-1}biggvert = lim_{x to 0}biggvertfrac{5^{2x}-1}{5^{2x}-5^x}biggvert = lim_{x to 0}biggvertfrac{(5^{x}-1)(5^x+1)}{5^{x}(5^x-1)}biggvert$$
answered Nov 30 '18 at 12:33
KM101KM101
5,9431523
$endgroup$
add a comment |
$begingroup$
Set $y= 5^x$ and consider $ylongrightarrow 1$
$$left|frac{5^x - 5^{-x}}{5^x-1}right| = left|frac{y-frac{1}{y}}{y-1}right| = left|frac{y^2-1}{y(y-1)}right| = left|frac{y+1}{y}right| stackrel{y to 1}{longrightarrow} 2$$
answered Nov 30 '18 at 12:32
trancelocationtrancelocation
9,8101722
$endgroup$
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$begingroup$
HINT
For positive $x$ we have:
$frac{5^x-5^{-x}}{5^x-1}= frac{(5^x-1)+(1-5^{-x})}{5^x-1}= 1+5^{-x}$.
As $xrightarrow 0$ this approaches to $1+1=2$.
answered Nov 30 '18 at 12:52
A. PongráczA. Pongrácz
5,9831929
$endgroup$
add a comment |
$begingroup$
Alternative approach: Recall that $lim_{xto0}frac{a^x-1}{x}=ln a$
$lim_{xto 0}left|frac{5^x - 5^{-x}}{5^x-1}right|=lim_{xto 0}left|frac{5^{2x}-1 }{5^x(5^x-1)}right|=lim_{xto 0} |frac{5^{2x}-1 }{2x}|cdotfrac{2}{5^x}cdot|frac{x}{5^x-1}|=frac{2ln5}{ln5}=2 $
answered Nov 30 '18 at 12:36
Shubham JohriShubham Johri
4,785717
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
$$lim_{x to 0}biggvertfrac{5^x-5^{-x}}{5^x-1}biggvert = lim_{x to 0}biggvertfrac{5^{2x}-1}{5^{2x}-5^x}biggvert = lim_{x to 0}biggvertfrac{(5^{x}-1)(5^x+1)}{5^{x}(5^x-1)}biggvert$$
answered Nov 30 '18 at 12:33
KM101KM101
5,9431523
$endgroup$
add a comment |
$begingroup$
Hint:
$$lim_{x to 0}biggvertfrac{5^x-5^{-x}}{5^x-1}biggvert = lim_{x to 0}biggvertfrac{5^{2x}-1}{5^{2x}-5^x}biggvert = lim_{x to 0}biggvertfrac{(5^{x}-1)(5^x+1)}{5^{x}(5^x-1)}biggvert$$
answered Nov 30 '18 at 12:33
KM101KM101
5,9431523
$endgroup$
add a comment |
$begingroup$
Hint:
$$lim_{x to 0}biggvertfrac{5^x-5^{-x}}{5^x-1}biggvert = lim_{x to 0}biggvertfrac{5^{2x}-1}{5^{2x}-5^x}biggvert = lim_{x to 0}biggvertfrac{(5^{x}-1)(5^x+1)}{5^{x}(5^x-1)}biggvert$$
answered Nov 30 '18 at 12:33
KM101KM101
5,9431523
$endgroup$
Hint:
$$lim_{x to 0}biggvertfrac{5^x-5^{-x}}{5^x-1}biggvert = lim_{x to 0}biggvertfrac{5^{2x}-1}{5^{2x}-5^x}biggvert = lim_{x to 0}biggvertfrac{(5^{x}-1)(5^x+1)}{5^{x}(5^x-1)}biggvert$$
answered Nov 30 '18 at 12:33
KM101KM101
5,9431523
answered Nov 30 '18 at 12:33
KM101KM101
5,9431523
answered Nov 30 '18 at 12:33
KM101KM101
5,9431523
answered Nov 30 '18 at 12:33
KM101KM101
5,9431523
5,9431523
add a comment |
add a comment |
$begingroup$
Set $y= 5^x$ and consider $ylongrightarrow 1$
$$left|frac{5^x - 5^{-x}}{5^x-1}right| = left|frac{y-frac{1}{y}}{y-1}right| = left|frac{y^2-1}{y(y-1)}right| = left|frac{y+1}{y}right| stackrel{y to 1}{longrightarrow} 2$$
answered Nov 30 '18 at 12:32
trancelocationtrancelocation
9,8101722
$endgroup$
add a comment |
$begingroup$
Set $y= 5^x$ and consider $ylongrightarrow 1$
$$left|frac{5^x - 5^{-x}}{5^x-1}right| = left|frac{y-frac{1}{y}}{y-1}right| = left|frac{y^2-1}{y(y-1)}right| = left|frac{y+1}{y}right| stackrel{y to 1}{longrightarrow} 2$$
answered Nov 30 '18 at 12:32
trancelocationtrancelocation
9,8101722
$endgroup$
add a comment |
$begingroup$
Set $y= 5^x$ and consider $ylongrightarrow 1$
$$left|frac{5^x - 5^{-x}}{5^x-1}right| = left|frac{y-frac{1}{y}}{y-1}right| = left|frac{y^2-1}{y(y-1)}right| = left|frac{y+1}{y}right| stackrel{y to 1}{longrightarrow} 2$$
answered Nov 30 '18 at 12:32
trancelocationtrancelocation
9,8101722
$endgroup$
Set $y= 5^x$ and consider $ylongrightarrow 1$
$$left|frac{5^x - 5^{-x}}{5^x-1}right| = left|frac{y-frac{1}{y}}{y-1}right| = left|frac{y^2-1}{y(y-1)}right| = left|frac{y+1}{y}right| stackrel{y to 1}{longrightarrow} 2$$
answered Nov 30 '18 at 12:32
trancelocationtrancelocation
9,8101722
answered Nov 30 '18 at 12:32
trancelocationtrancelocation
9,8101722
answered Nov 30 '18 at 12:32
trancelocationtrancelocation
9,8101722
answered Nov 30 '18 at 12:32
trancelocationtrancelocation
9,8101722
9,8101722
add a comment |
add a comment |
$begingroup$
HINT
For positive $x$ we have:
$frac{5^x-5^{-x}}{5^x-1}= frac{(5^x-1)+(1-5^{-x})}{5^x-1}= 1+5^{-x}$.
As $xrightarrow 0$ this approaches to $1+1=2$.
answered Nov 30 '18 at 12:52
A. PongráczA. Pongrácz
5,9831929
$endgroup$
add a comment |
$begingroup$
HINT
For positive $x$ we have:
$frac{5^x-5^{-x}}{5^x-1}= frac{(5^x-1)+(1-5^{-x})}{5^x-1}= 1+5^{-x}$.
As $xrightarrow 0$ this approaches to $1+1=2$.
answered Nov 30 '18 at 12:52
A. PongráczA. Pongrácz
5,9831929
$endgroup$
add a comment |
$begingroup$
HINT
For positive $x$ we have:
$frac{5^x-5^{-x}}{5^x-1}= frac{(5^x-1)+(1-5^{-x})}{5^x-1}= 1+5^{-x}$.
As $xrightarrow 0$ this approaches to $1+1=2$.
answered Nov 30 '18 at 12:52
A. PongráczA. Pongrácz
5,9831929
$endgroup$
HINT
For positive $x$ we have:
$frac{5^x-5^{-x}}{5^x-1}= frac{(5^x-1)+(1-5^{-x})}{5^x-1}= 1+5^{-x}$.
As $xrightarrow 0$ this approaches to $1+1=2$.
answered Nov 30 '18 at 12:52
A. PongráczA. Pongrácz
5,9831929
answered Nov 30 '18 at 12:52
A. PongráczA. Pongrácz
5,9831929
answered Nov 30 '18 at 12:52
A. PongráczA. Pongrácz
5,9831929
answered Nov 30 '18 at 12:52
A. PongráczA. Pongrácz
5,9831929
5,9831929
add a comment |
add a comment |
$begingroup$
Alternative approach: Recall that $lim_{xto0}frac{a^x-1}{x}=ln a$
$lim_{xto 0}left|frac{5^x - 5^{-x}}{5^x-1}right|=lim_{xto 0}left|frac{5^{2x}-1 }{5^x(5^x-1)}right|=lim_{xto 0} |frac{5^{2x}-1 }{2x}|cdotfrac{2}{5^x}cdot|frac{x}{5^x-1}|=frac{2ln5}{ln5}=2 $
answered Nov 30 '18 at 12:36
Shubham JohriShubham Johri
4,785717
$endgroup$
add a comment |
$begingroup$
Alternative approach: Recall that $lim_{xto0}frac{a^x-1}{x}=ln a$
$lim_{xto 0}left|frac{5^x - 5^{-x}}{5^x-1}right|=lim_{xto 0}left|frac{5^{2x}-1 }{5^x(5^x-1)}right|=lim_{xto 0} |frac{5^{2x}-1 }{2x}|cdotfrac{2}{5^x}cdot|frac{x}{5^x-1}|=frac{2ln5}{ln5}=2 $
answered Nov 30 '18 at 12:36
Shubham JohriShubham Johri
4,785717
$endgroup$
add a comment |
$begingroup$
Alternative approach: Recall that $lim_{xto0}frac{a^x-1}{x}=ln a$
$lim_{xto 0}left|frac{5^x - 5^{-x}}{5^x-1}right|=lim_{xto 0}left|frac{5^{2x}-1 }{5^x(5^x-1)}right|=lim_{xto 0} |frac{5^{2x}-1 }{2x}|cdotfrac{2}{5^x}cdot|frac{x}{5^x-1}|=frac{2ln5}{ln5}=2 $
answered Nov 30 '18 at 12:36
Shubham JohriShubham Johri
4,785717
$endgroup$
Alternative approach: Recall that $lim_{xto0}frac{a^x-1}{x}=ln a$
$lim_{xto 0}left|frac{5^x - 5^{-x}}{5^x-1}right|=lim_{xto 0}left|frac{5^{2x}-1 }{5^x(5^x-1)}right|=lim_{xto 0} |frac{5^{2x}-1 }{2x}|cdotfrac{2}{5^x}cdot|frac{x}{5^x-1}|=frac{2ln5}{ln5}=2 $
answered Nov 30 '18 at 12:36
Shubham JohriShubham Johri
4,785717
answered Nov 30 '18 at 12:36
Shubham JohriShubham Johri
4,785717
answered Nov 30 '18 at 12:36
Shubham JohriShubham Johri
4,785717
answered Nov 30 '18 at 12:36
Shubham JohriShubham Johri
4,785717
4,785717
add a comment |
add a comment |
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