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Calculating limit of function composition
The Next CEO of Stack Overflowfinding a limit of a function by definitionProve or disprove the existence of a limit of the function $displaystyle lim_xto x_0 f(x)$Prove left limit equals right limitHow to prove that a divergent limit of a function multiplied by a convergent limit of a function is divergentCan I write: $lim limits _x to x_0 (f(x) + g(x)) = lim limits _x to x_0 f(x) + lim limits _x to x_0 g(x)$?How to prove $limlimits_nto infty(a_n+2b_n)=L+2M$ using the formal definition of limit of sequence?How do you prove that if a function has a limit, its square has a limit, without using properties?Why is the limit of a real sequence defined so?A question about if and if only condition of limitWhat's the definition of this limit?
$begingroup$
Suppose we know that
$$
exists lim(h(x))_x to x_0 - = p in Bbb R
$$
$$
exists lim(s(t))_t to x_0- = x_0
$$
Where $h(x), s(t)$ - functions with domain $D=(a,x_0)$. What is more, $s(t) in D ;; forall t in D$. Can we assert the following:
$$
exists lim(h(s(t)))_t to x_0- = p
$$
How to prove/disprove it rigorously in terms of $(ε, δ)$-definition of limits?
calculus limits
asked Mar 27 at 18:38
Yan KardziyakaYan Kardziyaka
103
$endgroup$
add a comment |
$begingroup$
Suppose we know that
$$
exists lim(h(x))_x to x_0 - = p in Bbb R
$$
$$
exists lim(s(t))_t to x_0- = x_0
$$
Where $h(x), s(t)$ - functions with domain $D=(a,x_0)$. What is more, $s(t) in D ;; forall t in D$. Can we assert the following:
$$
exists lim(h(s(t)))_t to x_0- = p
$$
How to prove/disprove it rigorously in terms of $(ε, δ)$-definition of limits?
calculus limits
asked Mar 27 at 18:38
Yan KardziyakaYan Kardziyaka
103
$endgroup$
$begingroup$
Hint: use the $delta$ from the limit for $h$ as the $epsilon$ in the limit for $s$.
$endgroup$
– eyeballfrog
Mar 27 at 19:09
add a comment |
$begingroup$
Suppose we know that
$$
exists lim(h(x))_x to x_0 - = p in Bbb R
$$
$$
exists lim(s(t))_t to x_0- = x_0
$$
Where $h(x), s(t)$ - functions with domain $D=(a,x_0)$. What is more, $s(t) in D ;; forall t in D$. Can we assert the following:
$$
exists lim(h(s(t)))_t to x_0- = p
$$
How to prove/disprove it rigorously in terms of $(ε, δ)$-definition of limits?
calculus limits
asked Mar 27 at 18:38
Yan KardziyakaYan Kardziyaka
103
$endgroup$
Suppose we know that
$$
exists lim(h(x))_x to x_0 - = p in Bbb R
$$
$$
exists lim(s(t))_t to x_0- = x_0
$$
Where $h(x), s(t)$ - functions with domain $D=(a,x_0)$. What is more, $s(t) in D ;; forall t in D$. Can we assert the following:
$$
exists lim(h(s(t)))_t to x_0- = p
$$
How to prove/disprove it rigorously in terms of $(ε, δ)$-definition of limits?
calculus limits
calculus limits
asked Mar 27 at 18:38
Yan KardziyakaYan Kardziyaka
103
asked Mar 27 at 18:38
Yan KardziyakaYan Kardziyaka
103
asked Mar 27 at 18:38
Yan KardziyakaYan Kardziyaka
103
asked Mar 27 at 18:38
Yan KardziyakaYan Kardziyaka
103
asked Mar 27 at 18:38
Yan KardziyakaYan Kardziyaka
103
103
$begingroup$
Hint: use the $delta$ from the limit for $h$ as the $epsilon$ in the limit for $s$.
$endgroup$
– eyeballfrog
Mar 27 at 19:09
add a comment |
$begingroup$
Hint: use the $delta$ from the limit for $h$ as the $epsilon$ in the limit for $s$.
$endgroup$
– eyeballfrog
Mar 27 at 19:09
$begingroup$
Hint: use the $delta$ from the limit for $h$ as the $epsilon$ in the limit for $s$.
$endgroup$
– eyeballfrog
Mar 27 at 19:09
$begingroup$
Hint: use the $delta$ from the limit for $h$ as the $epsilon$ in the limit for $s$.
$endgroup$
– eyeballfrog
Mar 27 at 19:09
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Fix $varepsilon>0$. By assumption there exists a $delta_1>0$ such that $|h(x)-p|<varepsilon$ for all $xin D$ with $0<x_0-x<delta_1$. There also exists a $delta_2>0$ such that $|s(x)-x_0|<delta_1$ for all $xin D$ with $0<x_0-x<delta_2$. However, because $s(x)in D$ for all $x$ this means that in fact $x_0-s(x)<delta_1$ for all $xin D$ with $0<x_0-x<delta_2$.
Thus for all $xin D$, such that $0<x_0-x<delta_2$, we have $|h(s(x))-p|<varepsilon$, because $0<x_0-s(x)<delta_1$.
answered Mar 27 at 19:38
K.PowerK.Power
3,630926
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Fix $varepsilon>0$. By assumption there exists a $delta_1>0$ such that $|h(x)-p|<varepsilon$ for all $xin D$ with $0<x_0-x<delta_1$. There also exists a $delta_2>0$ such that $|s(x)-x_0|<delta_1$ for all $xin D$ with $0<x_0-x<delta_2$. However, because $s(x)in D$ for all $x$ this means that in fact $x_0-s(x)<delta_1$ for all $xin D$ with $0<x_0-x<delta_2$.
Thus for all $xin D$, such that $0<x_0-x<delta_2$, we have $|h(s(x))-p|<varepsilon$, because $0<x_0-s(x)<delta_1$.
answered Mar 27 at 19:38
K.PowerK.Power
3,630926
$endgroup$
add a comment |
$begingroup$
Fix $varepsilon>0$. By assumption there exists a $delta_1>0$ such that $|h(x)-p|<varepsilon$ for all $xin D$ with $0<x_0-x<delta_1$. There also exists a $delta_2>0$ such that $|s(x)-x_0|<delta_1$ for all $xin D$ with $0<x_0-x<delta_2$. However, because $s(x)in D$ for all $x$ this means that in fact $x_0-s(x)<delta_1$ for all $xin D$ with $0<x_0-x<delta_2$.
Thus for all $xin D$, such that $0<x_0-x<delta_2$, we have $|h(s(x))-p|<varepsilon$, because $0<x_0-s(x)<delta_1$.
answered Mar 27 at 19:38
K.PowerK.Power
3,630926
$endgroup$
add a comment |
$begingroup$
Fix $varepsilon>0$. By assumption there exists a $delta_1>0$ such that $|h(x)-p|<varepsilon$ for all $xin D$ with $0<x_0-x<delta_1$. There also exists a $delta_2>0$ such that $|s(x)-x_0|<delta_1$ for all $xin D$ with $0<x_0-x<delta_2$. However, because $s(x)in D$ for all $x$ this means that in fact $x_0-s(x)<delta_1$ for all $xin D$ with $0<x_0-x<delta_2$.
Thus for all $xin D$, such that $0<x_0-x<delta_2$, we have $|h(s(x))-p|<varepsilon$, because $0<x_0-s(x)<delta_1$.
answered Mar 27 at 19:38
K.PowerK.Power
3,630926
$endgroup$
Fix $varepsilon>0$. By assumption there exists a $delta_1>0$ such that $|h(x)-p|<varepsilon$ for all $xin D$ with $0<x_0-x<delta_1$. There also exists a $delta_2>0$ such that $|s(x)-x_0|<delta_1$ for all $xin D$ with $0<x_0-x<delta_2$. However, because $s(x)in D$ for all $x$ this means that in fact $x_0-s(x)<delta_1$ for all $xin D$ with $0<x_0-x<delta_2$.
Thus for all $xin D$, such that $0<x_0-x<delta_2$, we have $|h(s(x))-p|<varepsilon$, because $0<x_0-s(x)<delta_1$.
answered Mar 27 at 19:38
K.PowerK.Power
3,630926
answered Mar 27 at 19:38
K.PowerK.Power
3,630926
answered Mar 27 at 19:38
K.PowerK.Power
3,630926
answered Mar 27 at 19:38
K.PowerK.Power
3,630926
3,630926
add a comment |
add a comment |
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$begingroup$
Hint: use the $delta$ from the limit for $h$ as the $epsilon$ in the limit for $s$.
$endgroup$
– eyeballfrog
Mar 27 at 19:09