Isn't This 0/0?

Calculus Level 5

$\large \displaystyle \lim_{x \to \frac\pi2} \dfrac{\sin x - (\sin x)^{\sin x}}{1-\sin x +\ln (\sin x)}= \ ?$

1 Limit does not exist 0 2

1 solution

Rohit Ner
Oct 8, 2015

The limit can be rewritten as
$\begin{aligned} \lim _{ x\rightarrow 1 }{ \frac { x-{ x }^{ x } }{ 1-x+\ln { x } } } &=\lim _{ x\rightarrow 1 }{ \frac { 1-{ x }^{ x }\left( 1+\ln { x } \right) }{ -1+\frac { 1 }{ x } } } \\&=\lim _{ x\rightarrow 1 }{ \frac { -{ x }^{ x }{ \left( 1+\ln { x } \right) }^{ 2 }-{ x }^{ x-1 } }{ -\frac { 1 }{ { x }^{ 2 } } } } \\&=\lim _{ x\rightarrow 1 }{ { x }^{ 2 } } \left( { x }^{ x }{ \left( 1+\ln { x } \right) }^{ 2 }+{ x }^{ x-1 } \right) \\&\Huge\color{#3D99F6}{=\boxed{2}} \end{aligned}$

Refer L'Hôpital's Rule

Ramez Hindi - 5 years, 8 months ago

Oh my god!L'Hopital is meant to make life easier. Is this the only way to solve it without using the rule?

Rohit Ner - 5 years, 8 months ago

i have a long solution without using L'Hopital Rule or Series expansion . i tried to add it here but i failed due to the $ sign

Ramez Hindi - 5 years, 8 months ago

Long but yeah from basics liked!!

rajdeep brahma - 3 years, 2 months ago

Good solution sir!!+1

rajdeep brahma - 3 years, 2 months ago

Isn't This 0/0?

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