Find $lim_{x to 0} (sin x)^{1/x} + (1/x)^{sin x}$
$begingroup$
Evaluate the limit $lim_{x to 0} (sin x)^{1/x} + (1/x)^{sin x}$
This limit is of the form $(0)^{infty} + {infty}^{0}$
I tried to calculate both limits individually, and got confused if the first term is an indeterminate form of not. If it is not, then this value is $0$ and second term is $1$.
But if first term is an indeterminate form then how would i calculate the limit?
calculus
asked Jan 3 at 14:06
MathsaddictMathsaddict
3669
$endgroup$
add a comment |
$begingroup$
Evaluate the limit $lim_{x to 0} (sin x)^{1/x} + (1/x)^{sin x}$
This limit is of the form $(0)^{infty} + {infty}^{0}$
I tried to calculate both limits individually, and got confused if the first term is an indeterminate form of not. If it is not, then this value is $0$ and second term is $1$.
But if first term is an indeterminate form then how would i calculate the limit?
calculus
asked Jan 3 at 14:06
MathsaddictMathsaddict
3669
$endgroup$
1
$begingroup$
Is $x$ meant to tend to $0$ rather than $infty$? Maybe $0^+$?
$endgroup$
– Olivier Moschetta
Jan 3 at 14:33
$begingroup$
I believe it should be $xto 0^+$ as pointed out by @OlivierMoschetta, otherwise the limit does not exist
$endgroup$
– roman
Jan 3 at 14:54
$begingroup$
Yea it's 0. Sorry for inconvenience.
$endgroup$
– Mathsaddict
Jan 3 at 15:04
add a comment |
$begingroup$
Evaluate the limit $lim_{x to 0} (sin x)^{1/x} + (1/x)^{sin x}$
This limit is of the form $(0)^{infty} + {infty}^{0}$
I tried to calculate both limits individually, and got confused if the first term is an indeterminate form of not. If it is not, then this value is $0$ and second term is $1$.
But if first term is an indeterminate form then how would i calculate the limit?
calculus
asked Jan 3 at 14:06
MathsaddictMathsaddict
3669
$endgroup$
Evaluate the limit $lim_{x to 0} (sin x)^{1/x} + (1/x)^{sin x}$
This limit is of the form $(0)^{infty} + {infty}^{0}$
I tried to calculate both limits individually, and got confused if the first term is an indeterminate form of not. If it is not, then this value is $0$ and second term is $1$.
But if first term is an indeterminate form then how would i calculate the limit?
calculus
calculus
asked Jan 3 at 14:06
MathsaddictMathsaddict
3669
asked Jan 3 at 14:06
MathsaddictMathsaddict
3669
edited Jan 3 at 14:54
Mathsaddict
asked Jan 3 at 14:06
MathsaddictMathsaddict
3669
asked Jan 3 at 14:06
MathsaddictMathsaddict
3669
asked Jan 3 at 14:06
MathsaddictMathsaddict
3669
3669
1
$begingroup$
Is $x$ meant to tend to $0$ rather than $infty$? Maybe $0^+$?
$endgroup$
– Olivier Moschetta
Jan 3 at 14:33
$begingroup$
I believe it should be $xto 0^+$ as pointed out by @OlivierMoschetta, otherwise the limit does not exist
$endgroup$
– roman
Jan 3 at 14:54
$begingroup$
Yea it's 0. Sorry for inconvenience.
$endgroup$
– Mathsaddict
Jan 3 at 15:04
add a comment |
1
$begingroup$
Is $x$ meant to tend to $0$ rather than $infty$? Maybe $0^+$?
$endgroup$
– Olivier Moschetta
Jan 3 at 14:33
$begingroup$
I believe it should be $xto 0^+$ as pointed out by @OlivierMoschetta, otherwise the limit does not exist
$endgroup$
– roman
Jan 3 at 14:54
$begingroup$
Yea it's 0. Sorry for inconvenience.
$endgroup$
– Mathsaddict
Jan 3 at 15:04
1
1
$begingroup$
Is $x$ meant to tend to $0$ rather than $infty$? Maybe $0^+$?
$endgroup$
– Olivier Moschetta
Jan 3 at 14:33
$begingroup$
Is $x$ meant to tend to $0$ rather than $infty$? Maybe $0^+$?
$endgroup$
– Olivier Moschetta
Jan 3 at 14:33
$begingroup$
I believe it should be $xto 0^+$ as pointed out by @OlivierMoschetta, otherwise the limit does not exist
$endgroup$
– roman
Jan 3 at 14:54
$begingroup$
I believe it should be $xto 0^+$ as pointed out by @OlivierMoschetta, otherwise the limit does not exist
$endgroup$
– roman
Jan 3 at 14:54
$begingroup$
Yea it's 0. Sorry for inconvenience.
$endgroup$
– Mathsaddict
Jan 3 at 15:04
$begingroup$
Yea it's 0. Sorry for inconvenience.
$endgroup$
– Mathsaddict
Jan 3 at 15:04
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I'm assuming $xrightarrow 0^+$ rather than $xrightarrowinfty$. The first limit is not an indeterminate form, but here is a direct calculation anyway.
Assume $0<x<1$ so that $1/x>1$. For every $ain ]0,1[$ we then have
$$0<a^{1/x}<a$$
(recall for instance that $a^3<a^2$ for such $a$), in particular if $a=sin(x)$:
$$0<sin(x)^{1/x}leqsin(x)$$
By the squeeze theorem we obtain
$$lim_{xrightarrow 0^+}sin(x)^{1/x}=0$$
answered Jan 3 at 14:43
Olivier MoschettaOlivier Moschetta
2,8311411
$endgroup$
$begingroup$
First term is not indeterminate. Why? When $x to 0$ $sin x to 0$ and $1/x to infty$ this should be $0^{infty}$ indeterminate form.
$endgroup$
– Mathsaddict
Jan 3 at 15:18
$begingroup$
You can adapt my calculations to prove that if $f(x)rightarrow 0$ and $g(x)rightarrowinfty$ then $f(x)^{g(x)}rightarrow 0$. Essentially $g(x)>1$ and $|f(x)|<1$ if $x$ is small enough so that $|f(x)|^{g(x)}leq |f(x)|rightarrow 0$.
$endgroup$
– Olivier Moschetta
Jan 3 at 15:25
add a comment |
$begingroup$
$$
L = lim_{xto 0^+} left((sin x)^{1over x} + left({1over x}right)^{sin x}right) \
= lim_{xto 0^+} frac{(sin x)^{1over x}cdot x^{sin x} + 1}{x^{sin x}}
$$
By $sin x sim x$ as $xto 0$:
$$
frac{(sin x)^{1over x}cdot x^{sin x} + 1}{x^{sin x}} sim frac{x^{1over x}x^x + 1}{x^x} = frac{x^{x+{1over x}}+ 1}{x^x}
$$
By the fact that $lim_{xto 0^+} x^x = 1$:
$$
begin{align}
lim_{xto 0^+} frac{x^{x+{1over x}}+ 1}{x^x} &= lim_{xto 0^+} left(x^{x+{1over x}} + 1right) \
&= 1+ lim_{xto 0^+} x^x cdot x^{1over x} \
&= 1 + lim_{xto 0^+}x^x cdot lim_{xto 0^+} sqrt[x]{x}\
&= 1 + 1cdot 0 = 1
end{align}
$$
answered Jan 3 at 14:55
romanroman
2,42721226
$endgroup$
add a comment |
Your Answer
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
I'm assuming $xrightarrow 0^+$ rather than $xrightarrowinfty$. The first limit is not an indeterminate form, but here is a direct calculation anyway.
Assume $0<x<1$ so that $1/x>1$. For every $ain ]0,1[$ we then have
$$0<a^{1/x}<a$$
(recall for instance that $a^3<a^2$ for such $a$), in particular if $a=sin(x)$:
$$0<sin(x)^{1/x}leqsin(x)$$
By the squeeze theorem we obtain
$$lim_{xrightarrow 0^+}sin(x)^{1/x}=0$$
answered Jan 3 at 14:43
Olivier MoschettaOlivier Moschetta
2,8311411
$endgroup$
$begingroup$
First term is not indeterminate. Why? When $x to 0$ $sin x to 0$ and $1/x to infty$ this should be $0^{infty}$ indeterminate form.
$endgroup$
– Mathsaddict
Jan 3 at 15:18
$begingroup$
You can adapt my calculations to prove that if $f(x)rightarrow 0$ and $g(x)rightarrowinfty$ then $f(x)^{g(x)}rightarrow 0$. Essentially $g(x)>1$ and $|f(x)|<1$ if $x$ is small enough so that $|f(x)|^{g(x)}leq |f(x)|rightarrow 0$.
$endgroup$
– Olivier Moschetta
Jan 3 at 15:25
add a comment |
$begingroup$
I'm assuming $xrightarrow 0^+$ rather than $xrightarrowinfty$. The first limit is not an indeterminate form, but here is a direct calculation anyway.
Assume $0<x<1$ so that $1/x>1$. For every $ain ]0,1[$ we then have
$$0<a^{1/x}<a$$
(recall for instance that $a^3<a^2$ for such $a$), in particular if $a=sin(x)$:
$$0<sin(x)^{1/x}leqsin(x)$$
By the squeeze theorem we obtain
$$lim_{xrightarrow 0^+}sin(x)^{1/x}=0$$
answered Jan 3 at 14:43
Olivier MoschettaOlivier Moschetta
2,8311411
$endgroup$
$begingroup$
First term is not indeterminate. Why? When $x to 0$ $sin x to 0$ and $1/x to infty$ this should be $0^{infty}$ indeterminate form.
$endgroup$
– Mathsaddict
Jan 3 at 15:18
$begingroup$
You can adapt my calculations to prove that if $f(x)rightarrow 0$ and $g(x)rightarrowinfty$ then $f(x)^{g(x)}rightarrow 0$. Essentially $g(x)>1$ and $|f(x)|<1$ if $x$ is small enough so that $|f(x)|^{g(x)}leq |f(x)|rightarrow 0$.
$endgroup$
– Olivier Moschetta
Jan 3 at 15:25
add a comment |
$begingroup$
I'm assuming $xrightarrow 0^+$ rather than $xrightarrowinfty$. The first limit is not an indeterminate form, but here is a direct calculation anyway.
Assume $0<x<1$ so that $1/x>1$. For every $ain ]0,1[$ we then have
$$0<a^{1/x}<a$$
(recall for instance that $a^3<a^2$ for such $a$), in particular if $a=sin(x)$:
$$0<sin(x)^{1/x}leqsin(x)$$
By the squeeze theorem we obtain
$$lim_{xrightarrow 0^+}sin(x)^{1/x}=0$$
answered Jan 3 at 14:43
Olivier MoschettaOlivier Moschetta
2,8311411
$endgroup$
I'm assuming $xrightarrow 0^+$ rather than $xrightarrowinfty$. The first limit is not an indeterminate form, but here is a direct calculation anyway.
Assume $0<x<1$ so that $1/x>1$. For every $ain ]0,1[$ we then have
$$0<a^{1/x}<a$$
(recall for instance that $a^3<a^2$ for such $a$), in particular if $a=sin(x)$:
$$0<sin(x)^{1/x}leqsin(x)$$
By the squeeze theorem we obtain
$$lim_{xrightarrow 0^+}sin(x)^{1/x}=0$$
answered Jan 3 at 14:43
Olivier MoschettaOlivier Moschetta
2,8311411
answered Jan 3 at 14:43
Olivier MoschettaOlivier Moschetta
2,8311411
answered Jan 3 at 14:43
Olivier MoschettaOlivier Moschetta
2,8311411
answered Jan 3 at 14:43
Olivier MoschettaOlivier Moschetta
2,8311411
2,8311411
$begingroup$
First term is not indeterminate. Why? When $x to 0$ $sin x to 0$ and $1/x to infty$ this should be $0^{infty}$ indeterminate form.
$endgroup$
– Mathsaddict
Jan 3 at 15:18
$begingroup$
You can adapt my calculations to prove that if $f(x)rightarrow 0$ and $g(x)rightarrowinfty$ then $f(x)^{g(x)}rightarrow 0$. Essentially $g(x)>1$ and $|f(x)|<1$ if $x$ is small enough so that $|f(x)|^{g(x)}leq |f(x)|rightarrow 0$.
$endgroup$
– Olivier Moschetta
Jan 3 at 15:25
add a comment |
$begingroup$
First term is not indeterminate. Why? When $x to 0$ $sin x to 0$ and $1/x to infty$ this should be $0^{infty}$ indeterminate form.
$endgroup$
– Mathsaddict
Jan 3 at 15:18
$begingroup$
You can adapt my calculations to prove that if $f(x)rightarrow 0$ and $g(x)rightarrowinfty$ then $f(x)^{g(x)}rightarrow 0$. Essentially $g(x)>1$ and $|f(x)|<1$ if $x$ is small enough so that $|f(x)|^{g(x)}leq |f(x)|rightarrow 0$.
$endgroup$
– Olivier Moschetta
Jan 3 at 15:25
$begingroup$
First term is not indeterminate. Why? When $x to 0$ $sin x to 0$ and $1/x to infty$ this should be $0^{infty}$ indeterminate form.
$endgroup$
– Mathsaddict
Jan 3 at 15:18
$begingroup$
First term is not indeterminate. Why? When $x to 0$ $sin x to 0$ and $1/x to infty$ this should be $0^{infty}$ indeterminate form.
$endgroup$
– Mathsaddict
Jan 3 at 15:18
$begingroup$
You can adapt my calculations to prove that if $f(x)rightarrow 0$ and $g(x)rightarrowinfty$ then $f(x)^{g(x)}rightarrow 0$. Essentially $g(x)>1$ and $|f(x)|<1$ if $x$ is small enough so that $|f(x)|^{g(x)}leq |f(x)|rightarrow 0$.
$endgroup$
– Olivier Moschetta
Jan 3 at 15:25
$begingroup$
You can adapt my calculations to prove that if $f(x)rightarrow 0$ and $g(x)rightarrowinfty$ then $f(x)^{g(x)}rightarrow 0$. Essentially $g(x)>1$ and $|f(x)|<1$ if $x$ is small enough so that $|f(x)|^{g(x)}leq |f(x)|rightarrow 0$.
$endgroup$
– Olivier Moschetta
Jan 3 at 15:25
add a comment |
$begingroup$
$$
L = lim_{xto 0^+} left((sin x)^{1over x} + left({1over x}right)^{sin x}right) \
= lim_{xto 0^+} frac{(sin x)^{1over x}cdot x^{sin x} + 1}{x^{sin x}}
$$
By $sin x sim x$ as $xto 0$:
$$
frac{(sin x)^{1over x}cdot x^{sin x} + 1}{x^{sin x}} sim frac{x^{1over x}x^x + 1}{x^x} = frac{x^{x+{1over x}}+ 1}{x^x}
$$
By the fact that $lim_{xto 0^+} x^x = 1$:
$$
begin{align}
lim_{xto 0^+} frac{x^{x+{1over x}}+ 1}{x^x} &= lim_{xto 0^+} left(x^{x+{1over x}} + 1right) \
&= 1+ lim_{xto 0^+} x^x cdot x^{1over x} \
&= 1 + lim_{xto 0^+}x^x cdot lim_{xto 0^+} sqrt[x]{x}\
&= 1 + 1cdot 0 = 1
end{align}
$$
answered Jan 3 at 14:55
romanroman
2,42721226
$endgroup$
add a comment |
$begingroup$
$$
L = lim_{xto 0^+} left((sin x)^{1over x} + left({1over x}right)^{sin x}right) \
= lim_{xto 0^+} frac{(sin x)^{1over x}cdot x^{sin x} + 1}{x^{sin x}}
$$
By $sin x sim x$ as $xto 0$:
$$
frac{(sin x)^{1over x}cdot x^{sin x} + 1}{x^{sin x}} sim frac{x^{1over x}x^x + 1}{x^x} = frac{x^{x+{1over x}}+ 1}{x^x}
$$
By the fact that $lim_{xto 0^+} x^x = 1$:
$$
begin{align}
lim_{xto 0^+} frac{x^{x+{1over x}}+ 1}{x^x} &= lim_{xto 0^+} left(x^{x+{1over x}} + 1right) \
&= 1+ lim_{xto 0^+} x^x cdot x^{1over x} \
&= 1 + lim_{xto 0^+}x^x cdot lim_{xto 0^+} sqrt[x]{x}\
&= 1 + 1cdot 0 = 1
end{align}
$$
answered Jan 3 at 14:55
romanroman
2,42721226
$endgroup$
add a comment |
$begingroup$
$$
L = lim_{xto 0^+} left((sin x)^{1over x} + left({1over x}right)^{sin x}right) \
= lim_{xto 0^+} frac{(sin x)^{1over x}cdot x^{sin x} + 1}{x^{sin x}}
$$
By $sin x sim x$ as $xto 0$:
$$
frac{(sin x)^{1over x}cdot x^{sin x} + 1}{x^{sin x}} sim frac{x^{1over x}x^x + 1}{x^x} = frac{x^{x+{1over x}}+ 1}{x^x}
$$
By the fact that $lim_{xto 0^+} x^x = 1$:
$$
begin{align}
lim_{xto 0^+} frac{x^{x+{1over x}}+ 1}{x^x} &= lim_{xto 0^+} left(x^{x+{1over x}} + 1right) \
&= 1+ lim_{xto 0^+} x^x cdot x^{1over x} \
&= 1 + lim_{xto 0^+}x^x cdot lim_{xto 0^+} sqrt[x]{x}\
&= 1 + 1cdot 0 = 1
end{align}
$$
answered Jan 3 at 14:55
romanroman
2,42721226
$endgroup$
$$
L = lim_{xto 0^+} left((sin x)^{1over x} + left({1over x}right)^{sin x}right) \
= lim_{xto 0^+} frac{(sin x)^{1over x}cdot x^{sin x} + 1}{x^{sin x}}
$$
By $sin x sim x$ as $xto 0$:
$$
frac{(sin x)^{1over x}cdot x^{sin x} + 1}{x^{sin x}} sim frac{x^{1over x}x^x + 1}{x^x} = frac{x^{x+{1over x}}+ 1}{x^x}
$$
By the fact that $lim_{xto 0^+} x^x = 1$:
$$
begin{align}
lim_{xto 0^+} frac{x^{x+{1over x}}+ 1}{x^x} &= lim_{xto 0^+} left(x^{x+{1over x}} + 1right) \
&= 1+ lim_{xto 0^+} x^x cdot x^{1over x} \
&= 1 + lim_{xto 0^+}x^x cdot lim_{xto 0^+} sqrt[x]{x}\
&= 1 + 1cdot 0 = 1
end{align}
$$
answered Jan 3 at 14:55
romanroman
2,42721226
answered Jan 3 at 14:55
romanroman
2,42721226
answered Jan 3 at 14:55
romanroman
2,42721226
answered Jan 3 at 14:55
romanroman
2,42721226
2,42721226
add a comment |
add a comment |
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1
$begingroup$
Is $x$ meant to tend to $0$ rather than $infty$? Maybe $0^+$?
$endgroup$
– Olivier Moschetta
Jan 3 at 14:33
$begingroup$
I believe it should be $xto 0^+$ as pointed out by @OlivierMoschetta, otherwise the limit does not exist
$endgroup$
– roman
Jan 3 at 14:54
$begingroup$
Yea it's 0. Sorry for inconvenience.
$endgroup$
– Mathsaddict
Jan 3 at 15:04