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How can I find a limit using equivalent functions and substitutions ( without applying L'Hospital) for the...
-1
$begingroup$
How can I find a limit using equivalent functions and substitutions ( without applying L'Hospital) for the following problem?
There you can see the expression, I need to find the limit for, as x tends to 0
$exp[(cos(sqrt x)-1)/x]$, square brackets for clarity.
limits-without-lhopital
edited Jan 23 at 12:47
Namaste
1
asked Jan 22 at 18:28
Ieva BrakmaneIeva Brakmane
163
$endgroup$
|
show 3 more comments
-1
$begingroup$
How can I find a limit using equivalent functions and substitutions ( without applying L'Hospital) for the following problem?
There you can see the expression, I need to find the limit for, as x tends to 0
$exp[(cos(sqrt x)-1)/x]$, square brackets for clarity.
limits-without-lhopital
edited Jan 23 at 12:47
Namaste
1
asked Jan 22 at 18:28
Ieva BrakmaneIeva Brakmane
163
$endgroup$
$begingroup$
Hint: try using the power reduction formula for sine.
$endgroup$
– Jordan Green
Jan 22 at 18:42
$begingroup$
Then it would be $$exp[(sinx)/x]$$ which is one? Could it be the solution of the problem be $$e^1$$?
$endgroup$
– Ieva Brakmane
Jan 22 at 18:54
$begingroup$
Not quite. Note that $frac{ cos(sqrt{x}) - 1}{x} = frac{-2 (sin(sqrt{x}))^2}{x} = -2 (frac{sin(sqrt{x})}{x})^2$.
$endgroup$
– Jordan Green
Jan 22 at 19:01
$begingroup$
now I simplified: $$e^{-frac{2sin ^2left(sqrt{x}right)}{x^2}}$$ Then It I would take the root of x in front of the expression it would fraction with -2 so the solution would be $$e^{-1}$$?
$endgroup$
– Ieva Brakmane
Jan 22 at 19:19
$begingroup$
Sorry, I should have typed $frac{-2(sin(sqrt{x})}{x} = -2 ( frac{sin(sqrt{x})}{sqrt{x}})^2$. What can you say about $frac{sin(sqrt{x})}{sqrt{x}}$ as $x$ tends to $0$ from the right?
$endgroup$
– Jordan Green
Jan 22 at 19:19
|
show 3 more comments
-1
-1
-1
$begingroup$
How can I find a limit using equivalent functions and substitutions ( without applying L'Hospital) for the following problem?
There you can see the expression, I need to find the limit for, as x tends to 0
$exp[(cos(sqrt x)-1)/x]$, square brackets for clarity.
limits-without-lhopital
edited Jan 23 at 12:47
Namaste
1
asked Jan 22 at 18:28
Ieva BrakmaneIeva Brakmane
163
$endgroup$
How can I find a limit using equivalent functions and substitutions ( without applying L'Hospital) for the following problem?
There you can see the expression, I need to find the limit for, as x tends to 0
$exp[(cos(sqrt x)-1)/x]$, square brackets for clarity.
limits-without-lhopital
limits-without-lhopital
edited Jan 23 at 12:47
Namaste
1
asked Jan 22 at 18:28
Ieva BrakmaneIeva Brakmane
163
edited Jan 23 at 12:47
Namaste
1
asked Jan 22 at 18:28
Ieva BrakmaneIeva Brakmane
163
edited Jan 23 at 12:47
Namaste
1
edited Jan 23 at 12:47
Namaste
1
edited Jan 23 at 12:47
Namaste
1
1
asked Jan 22 at 18:28
Ieva BrakmaneIeva Brakmane
163
asked Jan 22 at 18:28
Ieva BrakmaneIeva Brakmane
163
asked Jan 22 at 18:28
Ieva BrakmaneIeva Brakmane
163
163
$begingroup$
Hint: try using the power reduction formula for sine.
$endgroup$
– Jordan Green
Jan 22 at 18:42
$begingroup$
Then it would be $$exp[(sinx)/x]$$ which is one? Could it be the solution of the problem be $$e^1$$?
$endgroup$
– Ieva Brakmane
Jan 22 at 18:54
$begingroup$
Not quite. Note that $frac{ cos(sqrt{x}) - 1}{x} = frac{-2 (sin(sqrt{x}))^2}{x} = -2 (frac{sin(sqrt{x})}{x})^2$.
$endgroup$
– Jordan Green
Jan 22 at 19:01
$begingroup$
now I simplified: $$e^{-frac{2sin ^2left(sqrt{x}right)}{x^2}}$$ Then It I would take the root of x in front of the expression it would fraction with -2 so the solution would be $$e^{-1}$$?
$endgroup$
– Ieva Brakmane
Jan 22 at 19:19
$begingroup$
Sorry, I should have typed $frac{-2(sin(sqrt{x})}{x} = -2 ( frac{sin(sqrt{x})}{sqrt{x}})^2$. What can you say about $frac{sin(sqrt{x})}{sqrt{x}}$ as $x$ tends to $0$ from the right?
$endgroup$
– Jordan Green
Jan 22 at 19:19
|
show 3 more comments
$begingroup$
Hint: try using the power reduction formula for sine.
$endgroup$
– Jordan Green
Jan 22 at 18:42
$begingroup$
Then it would be $$exp[(sinx)/x]$$ which is one? Could it be the solution of the problem be $$e^1$$?
$endgroup$
– Ieva Brakmane
Jan 22 at 18:54
$begingroup$
Not quite. Note that $frac{ cos(sqrt{x}) - 1}{x} = frac{-2 (sin(sqrt{x}))^2}{x} = -2 (frac{sin(sqrt{x})}{x})^2$.
$endgroup$
– Jordan Green
Jan 22 at 19:01
$begingroup$
now I simplified: $$e^{-frac{2sin ^2left(sqrt{x}right)}{x^2}}$$ Then It I would take the root of x in front of the expression it would fraction with -2 so the solution would be $$e^{-1}$$?
$endgroup$
– Ieva Brakmane
Jan 22 at 19:19
$begingroup$
Sorry, I should have typed $frac{-2(sin(sqrt{x})}{x} = -2 ( frac{sin(sqrt{x})}{sqrt{x}})^2$. What can you say about $frac{sin(sqrt{x})}{sqrt{x}}$ as $x$ tends to $0$ from the right?
$endgroup$
– Jordan Green
Jan 22 at 19:19
$begingroup$
Hint: try using the power reduction formula for sine.
$endgroup$
– Jordan Green
Jan 22 at 18:42
$begingroup$
Hint: try using the power reduction formula for sine.
$endgroup$
– Jordan Green
Jan 22 at 18:42
$begingroup$
Then it would be $$exp[(sinx)/x]$$ which is one? Could it be the solution of the problem be $$e^1$$?
$endgroup$
– Ieva Brakmane
Jan 22 at 18:54
$begingroup$
Then it would be $$exp[(sinx)/x]$$ which is one? Could it be the solution of the problem be $$e^1$$?
$endgroup$
– Ieva Brakmane
Jan 22 at 18:54
$begingroup$
Not quite. Note that $frac{ cos(sqrt{x}) - 1}{x} = frac{-2 (sin(sqrt{x}))^2}{x} = -2 (frac{sin(sqrt{x})}{x})^2$.
$endgroup$
– Jordan Green
Jan 22 at 19:01
$begingroup$
Not quite. Note that $frac{ cos(sqrt{x}) - 1}{x} = frac{-2 (sin(sqrt{x}))^2}{x} = -2 (frac{sin(sqrt{x})}{x})^2$.
$endgroup$
– Jordan Green
Jan 22 at 19:01
$begingroup$
now I simplified: $$e^{-frac{2sin ^2left(sqrt{x}right)}{x^2}}$$ Then It I would take the root of x in front of the expression it would fraction with -2 so the solution would be $$e^{-1}$$?
$endgroup$
– Ieva Brakmane
Jan 22 at 19:19
$begingroup$
now I simplified: $$e^{-frac{2sin ^2left(sqrt{x}right)}{x^2}}$$ Then It I would take the root of x in front of the expression it would fraction with -2 so the solution would be $$e^{-1}$$?
$endgroup$
– Ieva Brakmane
Jan 22 at 19:19
$begingroup$
Sorry, I should have typed $frac{-2(sin(sqrt{x})}{x} = -2 ( frac{sin(sqrt{x})}{sqrt{x}})^2$. What can you say about $frac{sin(sqrt{x})}{sqrt{x}}$ as $x$ tends to $0$ from the right?
$endgroup$
– Jordan Green
Jan 22 at 19:19
$begingroup$
Sorry, I should have typed $frac{-2(sin(sqrt{x})}{x} = -2 ( frac{sin(sqrt{x})}{sqrt{x}})^2$. What can you say about $frac{sin(sqrt{x})}{sqrt{x}}$ as $x$ tends to $0$ from the right?
$endgroup$
– Jordan Green
Jan 22 at 19:19
|
show 3 more comments
1 Answer
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$begingroup$
You may set $x = t^2$ and use
$lim_{yto 0}frac{1-cos y}{y^2} = frac{1}{2}$ (easy to verify with L'Hospital).
begin{eqnarray*} e^{frac{cossqrt x -1}{x}}
& = & e^{frac{cos t -1}{t^2}}\
& stackrel{tto 0}{longrightarrow} & e^{-frac{1}{2}} =frac{1}{sqrt{e}}\
end{eqnarray*}
answered Jan 23 at 12:03
trancelocationtrancelocation
12.6k1826
$endgroup$
add a comment |
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0
$begingroup$
You may set $x = t^2$ and use
$lim_{yto 0}frac{1-cos y}{y^2} = frac{1}{2}$ (easy to verify with L'Hospital).
begin{eqnarray*} e^{frac{cossqrt x -1}{x}}
& = & e^{frac{cos t -1}{t^2}}\
& stackrel{tto 0}{longrightarrow} & e^{-frac{1}{2}} =frac{1}{sqrt{e}}\
end{eqnarray*}
answered Jan 23 at 12:03
trancelocationtrancelocation
12.6k1826
$endgroup$
add a comment |
0
$begingroup$
You may set $x = t^2$ and use
$lim_{yto 0}frac{1-cos y}{y^2} = frac{1}{2}$ (easy to verify with L'Hospital).
begin{eqnarray*} e^{frac{cossqrt x -1}{x}}
& = & e^{frac{cos t -1}{t^2}}\
& stackrel{tto 0}{longrightarrow} & e^{-frac{1}{2}} =frac{1}{sqrt{e}}\
end{eqnarray*}
answered Jan 23 at 12:03
trancelocationtrancelocation
12.6k1826
$endgroup$
add a comment |
0
0
0
$begingroup$
You may set $x = t^2$ and use
$lim_{yto 0}frac{1-cos y}{y^2} = frac{1}{2}$ (easy to verify with L'Hospital).
begin{eqnarray*} e^{frac{cossqrt x -1}{x}}
& = & e^{frac{cos t -1}{t^2}}\
& stackrel{tto 0}{longrightarrow} & e^{-frac{1}{2}} =frac{1}{sqrt{e}}\
end{eqnarray*}
answered Jan 23 at 12:03
trancelocationtrancelocation
12.6k1826
$endgroup$
You may set $x = t^2$ and use
$lim_{yto 0}frac{1-cos y}{y^2} = frac{1}{2}$ (easy to verify with L'Hospital).
begin{eqnarray*} e^{frac{cossqrt x -1}{x}}
& = & e^{frac{cos t -1}{t^2}}\
& stackrel{tto 0}{longrightarrow} & e^{-frac{1}{2}} =frac{1}{sqrt{e}}\
end{eqnarray*}
answered Jan 23 at 12:03
trancelocationtrancelocation
12.6k1826
answered Jan 23 at 12:03
trancelocationtrancelocation
12.6k1826
answered Jan 23 at 12:03
trancelocationtrancelocation
12.6k1826
answered Jan 23 at 12:03
trancelocationtrancelocation
12.6k1826
12.6k1826
add a comment |
add a comment |
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$begingroup$
Hint: try using the power reduction formula for sine.
$endgroup$
– Jordan Green
Jan 22 at 18:42
$begingroup$
Then it would be $$exp[(sinx)/x]$$ which is one? Could it be the solution of the problem be $$e^1$$?
$endgroup$
– Ieva Brakmane
Jan 22 at 18:54
$begingroup$
Not quite. Note that $frac{ cos(sqrt{x}) - 1}{x} = frac{-2 (sin(sqrt{x}))^2}{x} = -2 (frac{sin(sqrt{x})}{x})^2$.
$endgroup$
– Jordan Green
Jan 22 at 19:01
$begingroup$
now I simplified: $$e^{-frac{2sin ^2left(sqrt{x}right)}{x^2}}$$ Then It I would take the root of x in front of the expression it would fraction with -2 so the solution would be $$e^{-1}$$?
$endgroup$
– Ieva Brakmane
Jan 22 at 19:19
$begingroup$
Sorry, I should have typed $frac{-2(sin(sqrt{x})}{x} = -2 ( frac{sin(sqrt{x})}{sqrt{x}})^2$. What can you say about $frac{sin(sqrt{x})}{sqrt{x}}$ as $x$ tends to $0$ from the right?
$endgroup$
– Jordan Green
Jan 22 at 19:19