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 Miike012 Oct9-11 06:25 PM

limit question...

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)...???

 gb7nash Oct9-11 06:37 PM

Re: limit question...

Quote:
 Quote by Miike012 (Post 3549102) 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?

 Miike012 Oct9-11 06:44 PM

Re: limit question...

cos(x) approaches 1..

 gb7nash Oct9-11 06:52 PM

Re: limit question...

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.

 Miike012 Oct9-11 06:56 PM

Re: limit question...

Thank you.

 Miike012 Oct9-11 06:57 PM

Re: limit question...

I wasn't aware that you can't use that rule if the function actually approaches a value.

 gb7nash Oct9-11 07:01 PM

Re: limit question...

Quote:
 Quote by Miike012 (Post 3549181) 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.

 Miike012 Oct9-11 07:02 PM

Re: limit question...

that makes sense. Thank you.

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