Evaluate the following limit .Find simple limit with basic methodsFind the limit without using L'Hopital's ruleExistence of limit for some sequences implies existence of limitEvaluate the limit using only the following results$lim_n rightarrow infty frac1-(1-1/n)^41-(1-1/n)^3$Evaluate the following limit without L'HopitalLimit of an indeterminate form with a quadratic expression under square rootLimit of an integralFinding the limit of a sequence of integralsEvaluate the limit of the following sequence .
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Evaluate the following limit .
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Evaluate the following limit .
Find simple limit with basic methodsFind the limit without using L'Hopital's ruleExistence of limit for some sequences implies existence of limitEvaluate the limit using only the following results$lim_n rightarrow infty frac1-(1-1/n)^41-(1-1/n)^3$Evaluate the following limit without L'HopitalLimit of an indeterminate form with a quadratic expression under square rootLimit of an integralFinding the limit of a sequence of integralsEvaluate the limit of the following sequence .
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
$mathbf The Problem is:$
Find $$lim_k to infty int_0^1 e^- (k^2x^2/2) dx$$
Actually, I have been expanding $e^- (k^2x^2/2)$ to find out the limit, but I can't approach further. I think it's divergent .
For any help, thanks in advance !!!
real-analysis limits
asked 8 hours ago
Rabi Kumar ChakrabortyRabi Kumar Chakraborty
3822 silver badges10 bronze badges
$endgroup$
add a comment |
$begingroup$
$mathbf The Problem is:$
Find $$lim_k to infty int_0^1 e^- (k^2x^2/2) dx$$
Actually, I have been expanding $e^- (k^2x^2/2)$ to find out the limit, but I can't approach further. I think it's divergent .
For any help, thanks in advance !!!
real-analysis limits
asked 8 hours ago
Rabi Kumar ChakrabortyRabi Kumar Chakraborty
3822 silver badges10 bronze badges
$endgroup$
$begingroup$
#Peter Foreman, because the values are getting increased by increase of values of $k$ .
$endgroup$
– Rabi Kumar Chakraborty
8 hours ago
$begingroup$
Do not accept my answer. it was not correct.
$endgroup$
– hamam_Abdallah
7 hours ago
add a comment |
$begingroup$
$mathbf The Problem is:$
Find $$lim_k to infty int_0^1 e^- (k^2x^2/2) dx$$
Actually, I have been expanding $e^- (k^2x^2/2)$ to find out the limit, but I can't approach further. I think it's divergent .
For any help, thanks in advance !!!
real-analysis limits
asked 8 hours ago
Rabi Kumar ChakrabortyRabi Kumar Chakraborty
3822 silver badges10 bronze badges
$endgroup$
$mathbf The Problem is:$
Find $$lim_k to infty int_0^1 e^- (k^2x^2/2) dx$$
Actually, I have been expanding $e^- (k^2x^2/2)$ to find out the limit, but I can't approach further. I think it's divergent .
For any help, thanks in advance !!!
real-analysis limits
real-analysis limits
asked 8 hours ago
Rabi Kumar ChakrabortyRabi Kumar Chakraborty
3822 silver badges10 bronze badges
asked 8 hours ago
Rabi Kumar ChakrabortyRabi Kumar Chakraborty
3822 silver badges10 bronze badges
asked 8 hours ago
Rabi Kumar ChakrabortyRabi Kumar Chakraborty
3822 silver badges10 bronze badges
asked 8 hours ago
Rabi Kumar ChakrabortyRabi Kumar Chakraborty
3822 silver badges10 bronze badges
asked 8 hours ago
Rabi Kumar ChakrabortyRabi Kumar Chakraborty
3822 silver badges10 bronze badges
3822 silver badges10 bronze badges
$begingroup$
#Peter Foreman, because the values are getting increased by increase of values of $k$ .
$endgroup$
– Rabi Kumar Chakraborty
8 hours ago
$begingroup$
Do not accept my answer. it was not correct.
$endgroup$
– hamam_Abdallah
7 hours ago
add a comment |
$begingroup$
#Peter Foreman, because the values are getting increased by increase of values of $k$ .
$endgroup$
– Rabi Kumar Chakraborty
8 hours ago
$begingroup$
Do not accept my answer. it was not correct.
$endgroup$
– hamam_Abdallah
7 hours ago
$begingroup$
#Peter Foreman, because the values are getting increased by increase of values of $k$ .
$endgroup$
– Rabi Kumar Chakraborty
8 hours ago
$begingroup$
#Peter Foreman, because the values are getting increased by increase of values of $k$ .
$endgroup$
– Rabi Kumar Chakraborty
8 hours ago
$begingroup$
Do not accept my answer. it was not correct.
$endgroup$
– hamam_Abdallah
7 hours ago
$begingroup$
Do not accept my answer. it was not correct.
$endgroup$
– hamam_Abdallah
7 hours ago
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
You want $$lim_ktoinftyfrac1kint_0^k e^-y^2/2dy=0$$by squeezing viz. $$0lefrac1kint_0^k e^-y^2/2dylefrac1kint_0^infty e^-y^2/2dy=fracsqrtpi/2k.$$
answered 8 hours ago
J.G.J.G.
43.8k2 gold badges39 silver badges60 bronze badges
$endgroup$
add a comment |
$begingroup$
You can majorise the integrand on the interval $(0,1)$ by a constant function $ge 1$ and apply the dominated convergence theorem. Thus, you can pass the limit under the integration. This yields $int_0^1 lim_k to infty e^-frack^2 x^22 mathrmdx =0$.
answered 7 hours ago
AbstractAbstract
385 bronze badges
$endgroup$
add a comment |
$begingroup$
The sequence $$ f_k= left(frac1e^x^2/2right)^k^2$$ is bounded by $1$ on the interval $[0,1]$ and $f_krightarrow 0$ uniformly since $1/e^x^2/2$ is decreasing so it is minimum is $1/sqrte$ so
$$lim int frac1e^fracx^2k^22=int lim frac1e^fracx^2k^22=0$$
answered 7 hours ago
AmeryrAmeryr
9713 silver badges13 bronze badges
$endgroup$
add a comment |
$begingroup$
My answer was not correct but i cannot delete it from mobile. thanks in advance.
answered 8 hours ago
hamam_Abdallahhamam_Abdallah
39k2 gold badges16 silver badges34 bronze badges
$endgroup$
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
$begingroup$
You want $$lim_ktoinftyfrac1kint_0^k e^-y^2/2dy=0$$by squeezing viz. $$0lefrac1kint_0^k e^-y^2/2dylefrac1kint_0^infty e^-y^2/2dy=fracsqrtpi/2k.$$
answered 8 hours ago
J.G.J.G.
43.8k2 gold badges39 silver badges60 bronze badges
$endgroup$
add a comment |
$begingroup$
You want $$lim_ktoinftyfrac1kint_0^k e^-y^2/2dy=0$$by squeezing viz. $$0lefrac1kint_0^k e^-y^2/2dylefrac1kint_0^infty e^-y^2/2dy=fracsqrtpi/2k.$$
answered 8 hours ago
J.G.J.G.
43.8k2 gold badges39 silver badges60 bronze badges
$endgroup$
add a comment |
$begingroup$
You want $$lim_ktoinftyfrac1kint_0^k e^-y^2/2dy=0$$by squeezing viz. $$0lefrac1kint_0^k e^-y^2/2dylefrac1kint_0^infty e^-y^2/2dy=fracsqrtpi/2k.$$
answered 8 hours ago
J.G.J.G.
43.8k2 gold badges39 silver badges60 bronze badges
$endgroup$
You want $$lim_ktoinftyfrac1kint_0^k e^-y^2/2dy=0$$by squeezing viz. $$0lefrac1kint_0^k e^-y^2/2dylefrac1kint_0^infty e^-y^2/2dy=fracsqrtpi/2k.$$
answered 8 hours ago
J.G.J.G.
43.8k2 gold badges39 silver badges60 bronze badges
answered 8 hours ago
J.G.J.G.
43.8k2 gold badges39 silver badges60 bronze badges
answered 8 hours ago
J.G.J.G.
43.8k2 gold badges39 silver badges60 bronze badges
answered 8 hours ago
J.G.J.G.
43.8k2 gold badges39 silver badges60 bronze badges
43.8k2 gold badges39 silver badges60 bronze badges
add a comment |
add a comment |
$begingroup$
You can majorise the integrand on the interval $(0,1)$ by a constant function $ge 1$ and apply the dominated convergence theorem. Thus, you can pass the limit under the integration. This yields $int_0^1 lim_k to infty e^-frack^2 x^22 mathrmdx =0$.
answered 7 hours ago
AbstractAbstract
385 bronze badges
$endgroup$
add a comment |
$begingroup$
You can majorise the integrand on the interval $(0,1)$ by a constant function $ge 1$ and apply the dominated convergence theorem. Thus, you can pass the limit under the integration. This yields $int_0^1 lim_k to infty e^-frack^2 x^22 mathrmdx =0$.
answered 7 hours ago
AbstractAbstract
385 bronze badges
$endgroup$
add a comment |
$begingroup$
You can majorise the integrand on the interval $(0,1)$ by a constant function $ge 1$ and apply the dominated convergence theorem. Thus, you can pass the limit under the integration. This yields $int_0^1 lim_k to infty e^-frack^2 x^22 mathrmdx =0$.
answered 7 hours ago
AbstractAbstract
385 bronze badges
$endgroup$
You can majorise the integrand on the interval $(0,1)$ by a constant function $ge 1$ and apply the dominated convergence theorem. Thus, you can pass the limit under the integration. This yields $int_0^1 lim_k to infty e^-frack^2 x^22 mathrmdx =0$.
answered 7 hours ago
AbstractAbstract
385 bronze badges
answered 7 hours ago
AbstractAbstract
385 bronze badges
answered 7 hours ago
AbstractAbstract
385 bronze badges
answered 7 hours ago
AbstractAbstract
385 bronze badges
385 bronze badges
add a comment |
add a comment |
$begingroup$
The sequence $$ f_k= left(frac1e^x^2/2right)^k^2$$ is bounded by $1$ on the interval $[0,1]$ and $f_krightarrow 0$ uniformly since $1/e^x^2/2$ is decreasing so it is minimum is $1/sqrte$ so
$$lim int frac1e^fracx^2k^22=int lim frac1e^fracx^2k^22=0$$
answered 7 hours ago
AmeryrAmeryr
9713 silver badges13 bronze badges
$endgroup$
add a comment |
$begingroup$
The sequence $$ f_k= left(frac1e^x^2/2right)^k^2$$ is bounded by $1$ on the interval $[0,1]$ and $f_krightarrow 0$ uniformly since $1/e^x^2/2$ is decreasing so it is minimum is $1/sqrte$ so
$$lim int frac1e^fracx^2k^22=int lim frac1e^fracx^2k^22=0$$
answered 7 hours ago
AmeryrAmeryr
9713 silver badges13 bronze badges
$endgroup$
add a comment |
$begingroup$
The sequence $$ f_k= left(frac1e^x^2/2right)^k^2$$ is bounded by $1$ on the interval $[0,1]$ and $f_krightarrow 0$ uniformly since $1/e^x^2/2$ is decreasing so it is minimum is $1/sqrte$ so
$$lim int frac1e^fracx^2k^22=int lim frac1e^fracx^2k^22=0$$
answered 7 hours ago
AmeryrAmeryr
9713 silver badges13 bronze badges
$endgroup$
The sequence $$ f_k= left(frac1e^x^2/2right)^k^2$$ is bounded by $1$ on the interval $[0,1]$ and $f_krightarrow 0$ uniformly since $1/e^x^2/2$ is decreasing so it is minimum is $1/sqrte$ so
$$lim int frac1e^fracx^2k^22=int lim frac1e^fracx^2k^22=0$$
answered 7 hours ago
AmeryrAmeryr
9713 silver badges13 bronze badges
answered 7 hours ago
AmeryrAmeryr
9713 silver badges13 bronze badges
answered 7 hours ago
AmeryrAmeryr
9713 silver badges13 bronze badges
answered 7 hours ago
AmeryrAmeryr
9713 silver badges13 bronze badges
9713 silver badges13 bronze badges
add a comment |
add a comment |
$begingroup$
My answer was not correct but i cannot delete it from mobile. thanks in advance.
answered 8 hours ago
hamam_Abdallahhamam_Abdallah
39k2 gold badges16 silver badges34 bronze badges
$endgroup$
add a comment |
$begingroup$
My answer was not correct but i cannot delete it from mobile. thanks in advance.
answered 8 hours ago
hamam_Abdallahhamam_Abdallah
39k2 gold badges16 silver badges34 bronze badges
$endgroup$
add a comment |
$begingroup$
My answer was not correct but i cannot delete it from mobile. thanks in advance.
answered 8 hours ago
hamam_Abdallahhamam_Abdallah
39k2 gold badges16 silver badges34 bronze badges
$endgroup$
My answer was not correct but i cannot delete it from mobile. thanks in advance.
answered 8 hours ago
hamam_Abdallahhamam_Abdallah
39k2 gold badges16 silver badges34 bronze badges
edited 7 hours ago
answered 8 hours ago
hamam_Abdallahhamam_Abdallah
39k2 gold badges16 silver badges34 bronze badges
answered 8 hours ago
hamam_Abdallahhamam_Abdallah
39k2 gold badges16 silver badges34 bronze badges
answered 8 hours ago
hamam_Abdallahhamam_Abdallah
39k2 gold badges16 silver badges34 bronze badges
39k2 gold badges16 silver badges34 bronze badges
add a comment |
add a comment |
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$begingroup$
#Peter Foreman, because the values are getting increased by increase of values of $k$ .
$endgroup$
– Rabi Kumar Chakraborty
8 hours ago
$begingroup$
Do not accept my answer. it was not correct.
$endgroup$
– hamam_Abdallah
7 hours ago