Evaluating $lim_{xto3}frac{e^{-1over(3-x)^2}+e(4-3cos(x-3))^{1/5}-e^{sqrt {4-x}}}{sqrt {1-cos(x-3)}}$












1












$begingroup$


First of all, I make the substitution $x-3=t$ which gives me: $$lim_{tto0}frac{e^{-1over t^2}+e(4-3cos t)^{1/5}-e^{sqrt {1-t}}}{sqrt {1-cos t}}$$
The term $e^{-1over t^2}$ goes to $0$ as $t$ approaches $0$ so I ignore it and $e^{sqrt {1-t}}$ is just $e$, so I have to work on remaining two terms.

The term in the denominator becomes ${tover sqrt2}$ as we replace $cos t$ by $1-{t^2over2}$. Applying the same to remaining term yields $e(1+{3t^2over2})^{1/5}=e(1+{3t^2over10})$. So we have: $$lim_{tto0}{e+{e3t^2over10}-eover{toversqrt 2}}$$
Which should equal to $0$, but apparently the correct answer is ${-eoversqrt2}$. So where am I making a mistake?






calculus limits







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Depends, are you evaluating the limit from the right or left? Because as written, this limit does not exist. Also, the assertion you make with $e(1+{3t^2over2})^{1/5}=e(1+{3t^2over10})$ is not true.
    $endgroup$
    – Denis28
    Jan 13 at 23:11












  • $begingroup$
    @Denis28 am evaluating limit just for $xto3$ not left or right.There's a possibility that the answer $({-eoversqrt 2})$ is flawed,but how doesn't the limit exist?
    $endgroup$
    – Turan Nasibli
    Jan 13 at 23:16








  • 1




    $begingroup$
    when evaluating the limit as $xto3^+$ and $xto3^- $you obtain different answers, thus the limit does not exist.
    $endgroup$
    – Denis28
    Jan 13 at 23:17










  • $begingroup$
    Then,something must be wrong with question,i guess.
    $endgroup$
    – Turan Nasibli
    Jan 13 at 23:19






  • 1




    $begingroup$
    The first term in numerator can be ignored but not because it tends to $0$ but because the term divided by denominator tends to $0$. In effect one is splitting the given expression into a sum of two fractions where the first fraction uses $e^{-1/(3-x)^2}$ as numerator and this fraction tends to $0$. It is best to apply algebra of limits rather than ignore terms.
    $endgroup$
    – Paramanand Singh
    Jan 14 at 4:22


















1












$begingroup$


First of all, I make the substitution $x-3=t$ which gives me: $$lim_{tto0}frac{e^{-1over t^2}+e(4-3cos t)^{1/5}-e^{sqrt {1-t}}}{sqrt {1-cos t}}$$
The term $e^{-1over t^2}$ goes to $0$ as $t$ approaches $0$ so I ignore it and $e^{sqrt {1-t}}$ is just $e$, so I have to work on remaining two terms.

The term in the denominator becomes ${tover sqrt2}$ as we replace $cos t$ by $1-{t^2over2}$. Applying the same to remaining term yields $e(1+{3t^2over2})^{1/5}=e(1+{3t^2over10})$. So we have: $$lim_{tto0}{e+{e3t^2over10}-eover{toversqrt 2}}$$
Which should equal to $0$, but apparently the correct answer is ${-eoversqrt2}$. So where am I making a mistake?






calculus limits







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Depends, are you evaluating the limit from the right or left? Because as written, this limit does not exist. Also, the assertion you make with $e(1+{3t^2over2})^{1/5}=e(1+{3t^2over10})$ is not true.
    $endgroup$
    – Denis28
    Jan 13 at 23:11












  • $begingroup$
    @Denis28 am evaluating limit just for $xto3$ not left or right.There's a possibility that the answer $({-eoversqrt 2})$ is flawed,but how doesn't the limit exist?
    $endgroup$
    – Turan Nasibli
    Jan 13 at 23:16








  • 1




    $begingroup$
    when evaluating the limit as $xto3^+$ and $xto3^- $you obtain different answers, thus the limit does not exist.
    $endgroup$
    – Denis28
    Jan 13 at 23:17










  • $begingroup$
    Then,something must be wrong with question,i guess.
    $endgroup$
    – Turan Nasibli
    Jan 13 at 23:19






  • 1




    $begingroup$
    The first term in numerator can be ignored but not because it tends to $0$ but because the term divided by denominator tends to $0$. In effect one is splitting the given expression into a sum of two fractions where the first fraction uses $e^{-1/(3-x)^2}$ as numerator and this fraction tends to $0$. It is best to apply algebra of limits rather than ignore terms.
    $endgroup$
    – Paramanand Singh
    Jan 14 at 4:22
















1












1








1





$begingroup$


First of all, I make the substitution $x-3=t$ which gives me: $$lim_{tto0}frac{e^{-1over t^2}+e(4-3cos t)^{1/5}-e^{sqrt {1-t}}}{sqrt {1-cos t}}$$
The term $e^{-1over t^2}$ goes to $0$ as $t$ approaches $0$ so I ignore it and $e^{sqrt {1-t}}$ is just $e$, so I have to work on remaining two terms.

The term in the denominator becomes ${tover sqrt2}$ as we replace $cos t$ by $1-{t^2over2}$. Applying the same to remaining term yields $e(1+{3t^2over2})^{1/5}=e(1+{3t^2over10})$. So we have: $$lim_{tto0}{e+{e3t^2over10}-eover{toversqrt 2}}$$
Which should equal to $0$, but apparently the correct answer is ${-eoversqrt2}$. So where am I making a mistake?






calculus limits







share|cite|improve this question











$endgroup$




First of all, I make the substitution $x-3=t$ which gives me: $$lim_{tto0}frac{e^{-1over t^2}+e(4-3cos t)^{1/5}-e^{sqrt {1-t}}}{sqrt {1-cos t}}$$
The term $e^{-1over t^2}$ goes to $0$ as $t$ approaches $0$ so I ignore it and $e^{sqrt {1-t}}$ is just $e$, so I have to work on remaining two terms.

The term in the denominator becomes ${tover sqrt2}$ as we replace $cos t$ by $1-{t^2over2}$. Applying the same to remaining term yields $e(1+{3t^2over2})^{1/5}=e(1+{3t^2over10})$. So we have: $$lim_{tto0}{e+{e3t^2over10}-eover{toversqrt 2}}$$
Which should equal to $0$, but apparently the correct answer is ${-eoversqrt2}$. So where am I making a mistake?






calculus limits




calculus limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 13 at 23:04









Dando18

4,71741235




4,71741235










asked Jan 13 at 22:57









Turan NasibliTuran Nasibli

846




846








  • 2




    $begingroup$
    Depends, are you evaluating the limit from the right or left? Because as written, this limit does not exist. Also, the assertion you make with $e(1+{3t^2over2})^{1/5}=e(1+{3t^2over10})$ is not true.
    $endgroup$
    – Denis28
    Jan 13 at 23:11












  • $begingroup$
    @Denis28 am evaluating limit just for $xto3$ not left or right.There's a possibility that the answer $({-eoversqrt 2})$ is flawed,but how doesn't the limit exist?
    $endgroup$
    – Turan Nasibli
    Jan 13 at 23:16








  • 1




    $begingroup$
    when evaluating the limit as $xto3^+$ and $xto3^- $you obtain different answers, thus the limit does not exist.
    $endgroup$
    – Denis28
    Jan 13 at 23:17










  • $begingroup$
    Then,something must be wrong with question,i guess.
    $endgroup$
    – Turan Nasibli
    Jan 13 at 23:19






  • 1




    $begingroup$
    The first term in numerator can be ignored but not because it tends to $0$ but because the term divided by denominator tends to $0$. In effect one is splitting the given expression into a sum of two fractions where the first fraction uses $e^{-1/(3-x)^2}$ as numerator and this fraction tends to $0$. It is best to apply algebra of limits rather than ignore terms.
    $endgroup$
    – Paramanand Singh
    Jan 14 at 4:22
















  • 2




    $begingroup$
    Depends, are you evaluating the limit from the right or left? Because as written, this limit does not exist. Also, the assertion you make with $e(1+{3t^2over2})^{1/5}=e(1+{3t^2over10})$ is not true.
    $endgroup$
    – Denis28
    Jan 13 at 23:11












  • $begingroup$
    @Denis28 am evaluating limit just for $xto3$ not left or right.There's a possibility that the answer $({-eoversqrt 2})$ is flawed,but how doesn't the limit exist?
    $endgroup$
    – Turan Nasibli
    Jan 13 at 23:16








  • 1




    $begingroup$
    when evaluating the limit as $xto3^+$ and $xto3^- $you obtain different answers, thus the limit does not exist.
    $endgroup$
    – Denis28
    Jan 13 at 23:17










  • $begingroup$
    Then,something must be wrong with question,i guess.
    $endgroup$
    – Turan Nasibli
    Jan 13 at 23:19






  • 1




    $begingroup$
    The first term in numerator can be ignored but not because it tends to $0$ but because the term divided by denominator tends to $0$. In effect one is splitting the given expression into a sum of two fractions where the first fraction uses $e^{-1/(3-x)^2}$ as numerator and this fraction tends to $0$. It is best to apply algebra of limits rather than ignore terms.
    $endgroup$
    – Paramanand Singh
    Jan 14 at 4:22










2




2




$begingroup$
Depends, are you evaluating the limit from the right or left? Because as written, this limit does not exist. Also, the assertion you make with $e(1+{3t^2over2})^{1/5}=e(1+{3t^2over10})$ is not true.
$endgroup$
– Denis28
Jan 13 at 23:11






$begingroup$
Depends, are you evaluating the limit from the right or left? Because as written, this limit does not exist. Also, the assertion you make with $e(1+{3t^2over2})^{1/5}=e(1+{3t^2over10})$ is not true.
$endgroup$
– Denis28
Jan 13 at 23:11














$begingroup$
@Denis28 am evaluating limit just for $xto3$ not left or right.There's a possibility that the answer $({-eoversqrt 2})$ is flawed,but how doesn't the limit exist?
$endgroup$
– Turan Nasibli
Jan 13 at 23:16






$begingroup$
@Denis28 am evaluating limit just for $xto3$ not left or right.There's a possibility that the answer $({-eoversqrt 2})$ is flawed,but how doesn't the limit exist?
$endgroup$
– Turan Nasibli
Jan 13 at 23:16






1




1




$begingroup$
when evaluating the limit as $xto3^+$ and $xto3^- $you obtain different answers, thus the limit does not exist.
$endgroup$
– Denis28
Jan 13 at 23:17




$begingroup$
when evaluating the limit as $xto3^+$ and $xto3^- $you obtain different answers, thus the limit does not exist.
$endgroup$
– Denis28
Jan 13 at 23:17












$begingroup$
Then,something must be wrong with question,i guess.
$endgroup$
– Turan Nasibli
Jan 13 at 23:19




$begingroup$
Then,something must be wrong with question,i guess.
$endgroup$
– Turan Nasibli
Jan 13 at 23:19




1




1




$begingroup$
The first term in numerator can be ignored but not because it tends to $0$ but because the term divided by denominator tends to $0$. In effect one is splitting the given expression into a sum of two fractions where the first fraction uses $e^{-1/(3-x)^2}$ as numerator and this fraction tends to $0$. It is best to apply algebra of limits rather than ignore terms.
$endgroup$
– Paramanand Singh
Jan 14 at 4:22






$begingroup$
The first term in numerator can be ignored but not because it tends to $0$ but because the term divided by denominator tends to $0$. In effect one is splitting the given expression into a sum of two fractions where the first fraction uses $e^{-1/(3-x)^2}$ as numerator and this fraction tends to $0$. It is best to apply algebra of limits rather than ignore terms.
$endgroup$
– Paramanand Singh
Jan 14 at 4:22












1 Answer
1






active

oldest

votes


















3












$begingroup$

Considering the limit $$lim_{xto3^-}frac{e^{frac{-1}{(3-x)^2}}+e(4-3cos(x-3))^{1/5}-e^{sqrt {4-x}}}{sqrt {1-cos(x-3)}}to{small{begin{bmatrix}
&t=x-3&\
&tto0^-&
end{bmatrix}}}tolim_{tto0^-}frac{e^{-frac{1}{t^2}}+e(4-3cos t)^{1/5}-e^{sqrt {1-t}}}{sqrt {1-cos t}}$$



And, as you mentioned it is enough to consider the limit
$$lim_{tto0^-}frac{e(4-3cos t)^{1/5}-e^{sqrt {1-t}}}{sqrt {1-cos t}}.$$



Using Taylor Series, we have
$$begin{aligned}
&(4-3cos t)^{1/5}sim 1+frac{3 t^2}{10}+cdots \
&e^{sqrt{1-t}}sim e-frac{e t}{2}+cdots\
&cos(t)sim1-frac{t^2}{2}+cdots
end{aligned}$$
and substituting (and noting that the asymptotic terms $to0$ as $tto0$) we have
$$lim_{tto0^-}frac{e+frac{3et^2}{10}-e+frac{e t}{2}}{frac{vert tvert}{sqrt{2}}}=-lim_{tto0^-}frac{frac{3et}{10}+frac{e}{2}}{frac{1}{sqrt{2}}}=-frac{e}{sqrt{2}}.$$




It is clear, from this last expression, that since there is a $vert tvert$ in the denominator the limit will not exist because $xto3^+implies tto0^+implies vert tvert=t$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much.So what was missing on my work was using taylor expansions of $e^{sqrt(1-t)}$.But how could i know that the answer i obtained $(0)$ was incorrect and that i should have expanded $e^{sqrt(1-t)}$?
    $endgroup$
    – Turan Nasibli
    Jan 13 at 23:54






  • 1




    $begingroup$
    Generally, you should expand each term up to the same order, in this case order $2$. You did not do such thing with $exp(sqrt{1-t})$
    $endgroup$
    – Denis28
    Jan 13 at 23:59












  • $begingroup$
    I didn't know that,thanks a lot.
    $endgroup$
    – Turan Nasibli
    Jan 14 at 0:03












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

Considering the limit $$lim_{xto3^-}frac{e^{frac{-1}{(3-x)^2}}+e(4-3cos(x-3))^{1/5}-e^{sqrt {4-x}}}{sqrt {1-cos(x-3)}}to{small{begin{bmatrix}
&t=x-3&\
&tto0^-&
end{bmatrix}}}tolim_{tto0^-}frac{e^{-frac{1}{t^2}}+e(4-3cos t)^{1/5}-e^{sqrt {1-t}}}{sqrt {1-cos t}}$$



And, as you mentioned it is enough to consider the limit
$$lim_{tto0^-}frac{e(4-3cos t)^{1/5}-e^{sqrt {1-t}}}{sqrt {1-cos t}}.$$



Using Taylor Series, we have
$$begin{aligned}
&(4-3cos t)^{1/5}sim 1+frac{3 t^2}{10}+cdots \
&e^{sqrt{1-t}}sim e-frac{e t}{2}+cdots\
&cos(t)sim1-frac{t^2}{2}+cdots
end{aligned}$$
and substituting (and noting that the asymptotic terms $to0$ as $tto0$) we have
$$lim_{tto0^-}frac{e+frac{3et^2}{10}-e+frac{e t}{2}}{frac{vert tvert}{sqrt{2}}}=-lim_{tto0^-}frac{frac{3et}{10}+frac{e}{2}}{frac{1}{sqrt{2}}}=-frac{e}{sqrt{2}}.$$




It is clear, from this last expression, that since there is a $vert tvert$ in the denominator the limit will not exist because $xto3^+implies tto0^+implies vert tvert=t$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much.So what was missing on my work was using taylor expansions of $e^{sqrt(1-t)}$.But how could i know that the answer i obtained $(0)$ was incorrect and that i should have expanded $e^{sqrt(1-t)}$?
    $endgroup$
    – Turan Nasibli
    Jan 13 at 23:54






  • 1




    $begingroup$
    Generally, you should expand each term up to the same order, in this case order $2$. You did not do such thing with $exp(sqrt{1-t})$
    $endgroup$
    – Denis28
    Jan 13 at 23:59












  • $begingroup$
    I didn't know that,thanks a lot.
    $endgroup$
    – Turan Nasibli
    Jan 14 at 0:03
















3












$begingroup$

Considering the limit $$lim_{xto3^-}frac{e^{frac{-1}{(3-x)^2}}+e(4-3cos(x-3))^{1/5}-e^{sqrt {4-x}}}{sqrt {1-cos(x-3)}}to{small{begin{bmatrix}
&t=x-3&\
&tto0^-&
end{bmatrix}}}tolim_{tto0^-}frac{e^{-frac{1}{t^2}}+e(4-3cos t)^{1/5}-e^{sqrt {1-t}}}{sqrt {1-cos t}}$$



And, as you mentioned it is enough to consider the limit
$$lim_{tto0^-}frac{e(4-3cos t)^{1/5}-e^{sqrt {1-t}}}{sqrt {1-cos t}}.$$



Using Taylor Series, we have
$$begin{aligned}
&(4-3cos t)^{1/5}sim 1+frac{3 t^2}{10}+cdots \
&e^{sqrt{1-t}}sim e-frac{e t}{2}+cdots\
&cos(t)sim1-frac{t^2}{2}+cdots
end{aligned}$$
and substituting (and noting that the asymptotic terms $to0$ as $tto0$) we have
$$lim_{tto0^-}frac{e+frac{3et^2}{10}-e+frac{e t}{2}}{frac{vert tvert}{sqrt{2}}}=-lim_{tto0^-}frac{frac{3et}{10}+frac{e}{2}}{frac{1}{sqrt{2}}}=-frac{e}{sqrt{2}}.$$




It is clear, from this last expression, that since there is a $vert tvert$ in the denominator the limit will not exist because $xto3^+implies tto0^+implies vert tvert=t$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much.So what was missing on my work was using taylor expansions of $e^{sqrt(1-t)}$.But how could i know that the answer i obtained $(0)$ was incorrect and that i should have expanded $e^{sqrt(1-t)}$?
    $endgroup$
    – Turan Nasibli
    Jan 13 at 23:54






  • 1




    $begingroup$
    Generally, you should expand each term up to the same order, in this case order $2$. You did not do such thing with $exp(sqrt{1-t})$
    $endgroup$
    – Denis28
    Jan 13 at 23:59












  • $begingroup$
    I didn't know that,thanks a lot.
    $endgroup$
    – Turan Nasibli
    Jan 14 at 0:03














3












3








3





$begingroup$

Considering the limit $$lim_{xto3^-}frac{e^{frac{-1}{(3-x)^2}}+e(4-3cos(x-3))^{1/5}-e^{sqrt {4-x}}}{sqrt {1-cos(x-3)}}to{small{begin{bmatrix}
&t=x-3&\
&tto0^-&
end{bmatrix}}}tolim_{tto0^-}frac{e^{-frac{1}{t^2}}+e(4-3cos t)^{1/5}-e^{sqrt {1-t}}}{sqrt {1-cos t}}$$



And, as you mentioned it is enough to consider the limit
$$lim_{tto0^-}frac{e(4-3cos t)^{1/5}-e^{sqrt {1-t}}}{sqrt {1-cos t}}.$$



Using Taylor Series, we have
$$begin{aligned}
&(4-3cos t)^{1/5}sim 1+frac{3 t^2}{10}+cdots \
&e^{sqrt{1-t}}sim e-frac{e t}{2}+cdots\
&cos(t)sim1-frac{t^2}{2}+cdots
end{aligned}$$
and substituting (and noting that the asymptotic terms $to0$ as $tto0$) we have
$$lim_{tto0^-}frac{e+frac{3et^2}{10}-e+frac{e t}{2}}{frac{vert tvert}{sqrt{2}}}=-lim_{tto0^-}frac{frac{3et}{10}+frac{e}{2}}{frac{1}{sqrt{2}}}=-frac{e}{sqrt{2}}.$$




It is clear, from this last expression, that since there is a $vert tvert$ in the denominator the limit will not exist because $xto3^+implies tto0^+implies vert tvert=t$.






share|cite|improve this answer









$endgroup$



Considering the limit $$lim_{xto3^-}frac{e^{frac{-1}{(3-x)^2}}+e(4-3cos(x-3))^{1/5}-e^{sqrt {4-x}}}{sqrt {1-cos(x-3)}}to{small{begin{bmatrix}
&t=x-3&\
&tto0^-&
end{bmatrix}}}tolim_{tto0^-}frac{e^{-frac{1}{t^2}}+e(4-3cos t)^{1/5}-e^{sqrt {1-t}}}{sqrt {1-cos t}}$$



And, as you mentioned it is enough to consider the limit
$$lim_{tto0^-}frac{e(4-3cos t)^{1/5}-e^{sqrt {1-t}}}{sqrt {1-cos t}}.$$



Using Taylor Series, we have
$$begin{aligned}
&(4-3cos t)^{1/5}sim 1+frac{3 t^2}{10}+cdots \
&e^{sqrt{1-t}}sim e-frac{e t}{2}+cdots\
&cos(t)sim1-frac{t^2}{2}+cdots
end{aligned}$$
and substituting (and noting that the asymptotic terms $to0$ as $tto0$) we have
$$lim_{tto0^-}frac{e+frac{3et^2}{10}-e+frac{e t}{2}}{frac{vert tvert}{sqrt{2}}}=-lim_{tto0^-}frac{frac{3et}{10}+frac{e}{2}}{frac{1}{sqrt{2}}}=-frac{e}{sqrt{2}}.$$




It is clear, from this last expression, that since there is a $vert tvert$ in the denominator the limit will not exist because $xto3^+implies tto0^+implies vert tvert=t$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 13 at 23:42









Denis28Denis28

884318




884318












  • $begingroup$
    Thank you very much.So what was missing on my work was using taylor expansions of $e^{sqrt(1-t)}$.But how could i know that the answer i obtained $(0)$ was incorrect and that i should have expanded $e^{sqrt(1-t)}$?
    $endgroup$
    – Turan Nasibli
    Jan 13 at 23:54






  • 1




    $begingroup$
    Generally, you should expand each term up to the same order, in this case order $2$. You did not do such thing with $exp(sqrt{1-t})$
    $endgroup$
    – Denis28
    Jan 13 at 23:59












  • $begingroup$
    I didn't know that,thanks a lot.
    $endgroup$
    – Turan Nasibli
    Jan 14 at 0:03


















  • $begingroup$
    Thank you very much.So what was missing on my work was using taylor expansions of $e^{sqrt(1-t)}$.But how could i know that the answer i obtained $(0)$ was incorrect and that i should have expanded $e^{sqrt(1-t)}$?
    $endgroup$
    – Turan Nasibli
    Jan 13 at 23:54






  • 1




    $begingroup$
    Generally, you should expand each term up to the same order, in this case order $2$. You did not do such thing with $exp(sqrt{1-t})$
    $endgroup$
    – Denis28
    Jan 13 at 23:59












  • $begingroup$
    I didn't know that,thanks a lot.
    $endgroup$
    – Turan Nasibli
    Jan 14 at 0:03
















$begingroup$
Thank you very much.So what was missing on my work was using taylor expansions of $e^{sqrt(1-t)}$.But how could i know that the answer i obtained $(0)$ was incorrect and that i should have expanded $e^{sqrt(1-t)}$?
$endgroup$
– Turan Nasibli
Jan 13 at 23:54




$begingroup$
Thank you very much.So what was missing on my work was using taylor expansions of $e^{sqrt(1-t)}$.But how could i know that the answer i obtained $(0)$ was incorrect and that i should have expanded $e^{sqrt(1-t)}$?
$endgroup$
– Turan Nasibli
Jan 13 at 23:54




1




1




$begingroup$
Generally, you should expand each term up to the same order, in this case order $2$. You did not do such thing with $exp(sqrt{1-t})$
$endgroup$
– Denis28
Jan 13 at 23:59






$begingroup$
Generally, you should expand each term up to the same order, in this case order $2$. You did not do such thing with $exp(sqrt{1-t})$
$endgroup$
– Denis28
Jan 13 at 23:59














$begingroup$
I didn't know that,thanks a lot.
$endgroup$
– Turan Nasibli
Jan 14 at 0:03




$begingroup$
I didn't know that,thanks a lot.
$endgroup$
– Turan Nasibli
Jan 14 at 0:03


















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