$lim_{xto 0^+} (sin x)^ {tan x}$












3












$begingroup$


I have stumbled on this question in one of my problem sets from Cal I and I'm not sure how to proceed after the last step.



$$lim_{xto 0^+} (sin x)^ {tan x}$$



// Applying exponential rule $$x=e^{ln(x)}$$



$$lim_{xto 0^+} exp[,ln((sin x)^ {tan x}),]$$



// Using natural logarithm property to bring the exponent to the front



$$lim_{xto 0^+} exp[,(tan x)ln(sin x),]$$



// Using an algebra trick where : $$x= frac{1}{frac{1}{x}}$$



$$lim_{xto 0^+} expleft[,frac{ln(sin x)}{frac{1}{tan x}},right]$$



// After this step, we were taught to check for Hospital's rule, however in this case when you plug the value $0$ in the numerator, you get $ln(sin 0)$ which is equal to $ln(0)$. This is the part that is confusing me since $ln(0)$ is undefined.



// Could someone please explain to me why this whole limit is equal to $1$? Does it have to do with the fact that $x$ is approaching $0$ from the right side?






limits







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    So your limit is $dfrac{infty}{infty}$ and with l'Hospital $$limdfrac{lnsin x}{cot x}=-dfrac12limsin2x=0$$
    $endgroup$
    – Nosrati
    Dec 6 '18 at 4:02


















3












$begingroup$


I have stumbled on this question in one of my problem sets from Cal I and I'm not sure how to proceed after the last step.



$$lim_{xto 0^+} (sin x)^ {tan x}$$



// Applying exponential rule $$x=e^{ln(x)}$$



$$lim_{xto 0^+} exp[,ln((sin x)^ {tan x}),]$$



// Using natural logarithm property to bring the exponent to the front



$$lim_{xto 0^+} exp[,(tan x)ln(sin x),]$$



// Using an algebra trick where : $$x= frac{1}{frac{1}{x}}$$



$$lim_{xto 0^+} expleft[,frac{ln(sin x)}{frac{1}{tan x}},right]$$



// After this step, we were taught to check for Hospital's rule, however in this case when you plug the value $0$ in the numerator, you get $ln(sin 0)$ which is equal to $ln(0)$. This is the part that is confusing me since $ln(0)$ is undefined.



// Could someone please explain to me why this whole limit is equal to $1$? Does it have to do with the fact that $x$ is approaching $0$ from the right side?






limits







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    So your limit is $dfrac{infty}{infty}$ and with l'Hospital $$limdfrac{lnsin x}{cot x}=-dfrac12limsin2x=0$$
    $endgroup$
    – Nosrati
    Dec 6 '18 at 4:02
















3












3








3


1



$begingroup$


I have stumbled on this question in one of my problem sets from Cal I and I'm not sure how to proceed after the last step.



$$lim_{xto 0^+} (sin x)^ {tan x}$$



// Applying exponential rule $$x=e^{ln(x)}$$



$$lim_{xto 0^+} exp[,ln((sin x)^ {tan x}),]$$



// Using natural logarithm property to bring the exponent to the front



$$lim_{xto 0^+} exp[,(tan x)ln(sin x),]$$



// Using an algebra trick where : $$x= frac{1}{frac{1}{x}}$$



$$lim_{xto 0^+} expleft[,frac{ln(sin x)}{frac{1}{tan x}},right]$$



// After this step, we were taught to check for Hospital's rule, however in this case when you plug the value $0$ in the numerator, you get $ln(sin 0)$ which is equal to $ln(0)$. This is the part that is confusing me since $ln(0)$ is undefined.



// Could someone please explain to me why this whole limit is equal to $1$? Does it have to do with the fact that $x$ is approaching $0$ from the right side?






limits







share|cite|improve this question











$endgroup$




I have stumbled on this question in one of my problem sets from Cal I and I'm not sure how to proceed after the last step.



$$lim_{xto 0^+} (sin x)^ {tan x}$$



// Applying exponential rule $$x=e^{ln(x)}$$



$$lim_{xto 0^+} exp[,ln((sin x)^ {tan x}),]$$



// Using natural logarithm property to bring the exponent to the front



$$lim_{xto 0^+} exp[,(tan x)ln(sin x),]$$



// Using an algebra trick where : $$x= frac{1}{frac{1}{x}}$$



$$lim_{xto 0^+} expleft[,frac{ln(sin x)}{frac{1}{tan x}},right]$$



// After this step, we were taught to check for Hospital's rule, however in this case when you plug the value $0$ in the numerator, you get $ln(sin 0)$ which is equal to $ln(0)$. This is the part that is confusing me since $ln(0)$ is undefined.



// Could someone please explain to me why this whole limit is equal to $1$? Does it have to do with the fact that $x$ is approaching $0$ from the right side?






limits




limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 4:01









Larry

2,39131129




2,39131129










asked Dec 6 '18 at 3:55









DBlykDBlyk

204




204








  • 1




    $begingroup$
    So your limit is $dfrac{infty}{infty}$ and with l'Hospital $$limdfrac{lnsin x}{cot x}=-dfrac12limsin2x=0$$
    $endgroup$
    – Nosrati
    Dec 6 '18 at 4:02
















  • 1




    $begingroup$
    So your limit is $dfrac{infty}{infty}$ and with l'Hospital $$limdfrac{lnsin x}{cot x}=-dfrac12limsin2x=0$$
    $endgroup$
    – Nosrati
    Dec 6 '18 at 4:02










1




1




$begingroup$
So your limit is $dfrac{infty}{infty}$ and with l'Hospital $$limdfrac{lnsin x}{cot x}=-dfrac12limsin2x=0$$
$endgroup$
– Nosrati
Dec 6 '18 at 4:02






$begingroup$
So your limit is $dfrac{infty}{infty}$ and with l'Hospital $$limdfrac{lnsin x}{cot x}=-dfrac12limsin2x=0$$
$endgroup$
– Nosrati
Dec 6 '18 at 4:02












1 Answer
1






active

oldest

votes


















2












$begingroup$

Remember what L'Hospital's rule says: it's not a direct substitution of $x = 0$, it's the limit: what happens as $x$ approaches $0$, and specifically from the right (since we have $0^+$)?



Generally, where the limits exist and all the other nuances associated with the definition, the rule states



$$lim_{x to c} frac{f(x)}{g(x)} =lim_{x to c} frac{f'(x)}{g'(x)}$$



if the original limit is one of the indeterminate forms $0/0$ or $pm infty / pm infty$.



Now, in your problem, since the exponential function is continuous, we can say



$$lim_{xto 0^+} expleft[,frac{ln(sin x)}{frac{1}{tan x}},right] = expleft[,lim_{xto 0^+} left( frac{ln(sin x)}{frac{1}{tan x}},right) right]$$



So notice, as $x to 0$ from the right - not simply $x=0$ - $sin(x) to 0$. Since $ln(x) to -infty$ as $x to 0$ from the right, and we have $sin(x) to 0$, it follows that $ln(sin(x))to -infty$ as $x to 0^+$.



Further, as $x to 0^+$, $tan(x) to 0$ and thus $1/tan(x)to+infty$.



Thus, we can apply L'Hospital's rule since the limit as-is is an indeterminate form $pm infty / pm infty$. Differentiate the numerator and denominator (preferably after swapping $cot(x) = 1/tan(x)$ for clarity's sake) and try to find the limit as $x to 0^+$ after.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    So derivative of $ln{(sin{x})}=cot{x}$ and the derivative of $cot{x}=-csc^2{x}$. So as $xrightarrow 0$ the term becomes $e^0=1$Hence the limit is one.
    $endgroup$
    – Dhamnekar Winod
    Dec 6 '18 at 6:29













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1 Answer
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active

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active

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2












$begingroup$

Remember what L'Hospital's rule says: it's not a direct substitution of $x = 0$, it's the limit: what happens as $x$ approaches $0$, and specifically from the right (since we have $0^+$)?



Generally, where the limits exist and all the other nuances associated with the definition, the rule states



$$lim_{x to c} frac{f(x)}{g(x)} =lim_{x to c} frac{f'(x)}{g'(x)}$$



if the original limit is one of the indeterminate forms $0/0$ or $pm infty / pm infty$.



Now, in your problem, since the exponential function is continuous, we can say



$$lim_{xto 0^+} expleft[,frac{ln(sin x)}{frac{1}{tan x}},right] = expleft[,lim_{xto 0^+} left( frac{ln(sin x)}{frac{1}{tan x}},right) right]$$



So notice, as $x to 0$ from the right - not simply $x=0$ - $sin(x) to 0$. Since $ln(x) to -infty$ as $x to 0$ from the right, and we have $sin(x) to 0$, it follows that $ln(sin(x))to -infty$ as $x to 0^+$.



Further, as $x to 0^+$, $tan(x) to 0$ and thus $1/tan(x)to+infty$.



Thus, we can apply L'Hospital's rule since the limit as-is is an indeterminate form $pm infty / pm infty$. Differentiate the numerator and denominator (preferably after swapping $cot(x) = 1/tan(x)$ for clarity's sake) and try to find the limit as $x to 0^+$ after.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    So derivative of $ln{(sin{x})}=cot{x}$ and the derivative of $cot{x}=-csc^2{x}$. So as $xrightarrow 0$ the term becomes $e^0=1$Hence the limit is one.
    $endgroup$
    – Dhamnekar Winod
    Dec 6 '18 at 6:29


















2












$begingroup$

Remember what L'Hospital's rule says: it's not a direct substitution of $x = 0$, it's the limit: what happens as $x$ approaches $0$, and specifically from the right (since we have $0^+$)?



Generally, where the limits exist and all the other nuances associated with the definition, the rule states



$$lim_{x to c} frac{f(x)}{g(x)} =lim_{x to c} frac{f'(x)}{g'(x)}$$



if the original limit is one of the indeterminate forms $0/0$ or $pm infty / pm infty$.



Now, in your problem, since the exponential function is continuous, we can say



$$lim_{xto 0^+} expleft[,frac{ln(sin x)}{frac{1}{tan x}},right] = expleft[,lim_{xto 0^+} left( frac{ln(sin x)}{frac{1}{tan x}},right) right]$$



So notice, as $x to 0$ from the right - not simply $x=0$ - $sin(x) to 0$. Since $ln(x) to -infty$ as $x to 0$ from the right, and we have $sin(x) to 0$, it follows that $ln(sin(x))to -infty$ as $x to 0^+$.



Further, as $x to 0^+$, $tan(x) to 0$ and thus $1/tan(x)to+infty$.



Thus, we can apply L'Hospital's rule since the limit as-is is an indeterminate form $pm infty / pm infty$. Differentiate the numerator and denominator (preferably after swapping $cot(x) = 1/tan(x)$ for clarity's sake) and try to find the limit as $x to 0^+$ after.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    So derivative of $ln{(sin{x})}=cot{x}$ and the derivative of $cot{x}=-csc^2{x}$. So as $xrightarrow 0$ the term becomes $e^0=1$Hence the limit is one.
    $endgroup$
    – Dhamnekar Winod
    Dec 6 '18 at 6:29
















2












2








2





$begingroup$

Remember what L'Hospital's rule says: it's not a direct substitution of $x = 0$, it's the limit: what happens as $x$ approaches $0$, and specifically from the right (since we have $0^+$)?



Generally, where the limits exist and all the other nuances associated with the definition, the rule states



$$lim_{x to c} frac{f(x)}{g(x)} =lim_{x to c} frac{f'(x)}{g'(x)}$$



if the original limit is one of the indeterminate forms $0/0$ or $pm infty / pm infty$.



Now, in your problem, since the exponential function is continuous, we can say



$$lim_{xto 0^+} expleft[,frac{ln(sin x)}{frac{1}{tan x}},right] = expleft[,lim_{xto 0^+} left( frac{ln(sin x)}{frac{1}{tan x}},right) right]$$



So notice, as $x to 0$ from the right - not simply $x=0$ - $sin(x) to 0$. Since $ln(x) to -infty$ as $x to 0$ from the right, and we have $sin(x) to 0$, it follows that $ln(sin(x))to -infty$ as $x to 0^+$.



Further, as $x to 0^+$, $tan(x) to 0$ and thus $1/tan(x)to+infty$.



Thus, we can apply L'Hospital's rule since the limit as-is is an indeterminate form $pm infty / pm infty$. Differentiate the numerator and denominator (preferably after swapping $cot(x) = 1/tan(x)$ for clarity's sake) and try to find the limit as $x to 0^+$ after.






share|cite|improve this answer









$endgroup$



Remember what L'Hospital's rule says: it's not a direct substitution of $x = 0$, it's the limit: what happens as $x$ approaches $0$, and specifically from the right (since we have $0^+$)?



Generally, where the limits exist and all the other nuances associated with the definition, the rule states



$$lim_{x to c} frac{f(x)}{g(x)} =lim_{x to c} frac{f'(x)}{g'(x)}$$



if the original limit is one of the indeterminate forms $0/0$ or $pm infty / pm infty$.



Now, in your problem, since the exponential function is continuous, we can say



$$lim_{xto 0^+} expleft[,frac{ln(sin x)}{frac{1}{tan x}},right] = expleft[,lim_{xto 0^+} left( frac{ln(sin x)}{frac{1}{tan x}},right) right]$$



So notice, as $x to 0$ from the right - not simply $x=0$ - $sin(x) to 0$. Since $ln(x) to -infty$ as $x to 0$ from the right, and we have $sin(x) to 0$, it follows that $ln(sin(x))to -infty$ as $x to 0^+$.



Further, as $x to 0^+$, $tan(x) to 0$ and thus $1/tan(x)to+infty$.



Thus, we can apply L'Hospital's rule since the limit as-is is an indeterminate form $pm infty / pm infty$. Differentiate the numerator and denominator (preferably after swapping $cot(x) = 1/tan(x)$ for clarity's sake) and try to find the limit as $x to 0^+$ after.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 6 '18 at 4:05









Eevee TrainerEevee Trainer

5,7781936




5,7781936








  • 2




    $begingroup$
    So derivative of $ln{(sin{x})}=cot{x}$ and the derivative of $cot{x}=-csc^2{x}$. So as $xrightarrow 0$ the term becomes $e^0=1$Hence the limit is one.
    $endgroup$
    – Dhamnekar Winod
    Dec 6 '18 at 6:29
















  • 2




    $begingroup$
    So derivative of $ln{(sin{x})}=cot{x}$ and the derivative of $cot{x}=-csc^2{x}$. So as $xrightarrow 0$ the term becomes $e^0=1$Hence the limit is one.
    $endgroup$
    – Dhamnekar Winod
    Dec 6 '18 at 6:29










2




2




$begingroup$
So derivative of $ln{(sin{x})}=cot{x}$ and the derivative of $cot{x}=-csc^2{x}$. So as $xrightarrow 0$ the term becomes $e^0=1$Hence the limit is one.
$endgroup$
– Dhamnekar Winod
Dec 6 '18 at 6:29






$begingroup$
So derivative of $ln{(sin{x})}=cot{x}$ and the derivative of $cot{x}=-csc^2{x}$. So as $xrightarrow 0$ the term becomes $e^0=1$Hence the limit is one.
$endgroup$
– Dhamnekar Winod
Dec 6 '18 at 6:29




















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