Solving the Limit of sin(cos(x))/sec(x) as x->0

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The limit of sin(cos(x))/sec(x) as x approaches 0 is evaluated to be sin(1). The discussion clarifies that while sin(x)/x approaches 1 as x approaches 0, this identity cannot be applied when the argument of the sine function approaches a value other than 0. Specifically, as x approaches 0, cos(x) approaches 1, leading to the limit being transformed into lim_{u -> 1} sin(u)/u, which does not allow the direct application of the sine limit identity.

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Miike012
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Problem:

Lim sin(cos(x))/sec(x)
x -> 0

The answer in the book is sin(1).. which is obvious... but why do I not arrive at the same answer doing is this way...

Knowing as x approaches zero, sin(x)/x approaches 1...

( Sin(cos(x))/cos(x) )(cos^2(x)) = (1)(1) = 1 = / = sin(1)...?
 
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Miike012 said:
Knowing as x approaches zero, sin(x)/x approaches 1...

( Sin(cos(x))/cos(x) )(cos^2(x)) = (1)(1) = 1 = / = sin(1)...?

The bolded part is correct. However, what is cos(x) approaching as x approaches 0?
 
cos(x) approaches 1..
 
Ok, so since cos(x) goes to 1:

\lim_{x \to 0} \frac{\sin(\cos(x))}{\cos(x)}

is equivalent to:

\lim_{u \to 1} \frac{\sin(u)}{u}

We can't apply the sin(_)/_ rule here. The term inside of the sine approaches 1.
 
Thank you.
 
I wasn't aware that you can't use that rule if the function actually approaches a value.
 
Miike012 said:
I wasn't aware that you can't use that rule if the function actually approaches a value.

\lim_{x \to 0}\frac{\sin(x)}{x} = 1

x needs to approach 0. If it approaches any other value (or does not exist), you can't use this identity.
 
that makes sense. Thank you.
 

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