find the limit

$\displaystyle \lim_{x\to0} \frac{sin(cos x)}{sec x}$

I can just plug in zero but my teacher says that we can't use the direct substitution property with trig functions even if you can get the correct answer, which is lame. So how do I solve this? And yes its sin(cos x) not sin x(cos x)

Quote:

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Originally Posted by yoman360

$\displaystyle \lim_{x\to0} \frac{sin(cos x)}{sec x}$

I can just plug in zero but my teacher says that we can't use the direct substitution property with trig functions

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What's wrong with using direct substitution with trig functions?!?!

$\displaystyle \lim_{x\to0}\frac{\sin\left(\cos x\right)}{\sec x}=\frac{\sin\left(\cos 0\right)}{\sec 0}=\sin\left(1\right)$

Quote:

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Originally Posted by Chris L T521

What's wrong with using direct substitution with trig functions?!?!

$\displaystyle \lim_{x\to0}\frac{\sin\left(\cos x\right)}{\sec x}=\frac{\sin\left(\cos 0\right)}{\sec 0}=\sin\left(1\right)$

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Because the teacher says the definition for direct substitutution property can only be used for polynomial and rational functions.

Theorem: (Direct Substitution Property) Suppose f(x) is a polyno-
mial or a rational function and a is in the domain of f.

http://web.viu.ca/wattsv/math121/Ove...bstitution.pdf

He said that "using the direct substitutution property for trig functions won't be theromatically (if thats a word) correct."

Seriously someone needs to revise the direct substitutution property so that it will be "theromatically" correct for trig functions.

Quote:

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Originally Posted by yoman360

$\displaystyle \lim_{x\to0} \frac{sin(cos x)}{sec x}$

I can just plug in zero but my teacher says that we can't use the direct substitution property with trig functions even if you can get the correct answer, which is lame. So how do I solve this? And yes its sin(cos x) not sin x(cos x)

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$\displaystyle \lim_{x\rightarrow 0}\frac{\sin (\cos x)}{\sec x}=\lim_{x\rightarrow 0}\sin (\cos x)\cos x=\lim_{x\rightarrow 0}\sin (\cos x)\lim_{x\rightarrow 0}\cos x$
$\displaystyle =\sin (\cos \lim_{x\rightarrow 0}x)\cos 0=\sin (\cos 0)\cos 0=\sin 1$
Note that there is also another theorem:
if $\displaystyle f(x)$is continuous at $\displaystyle x_0$, then
$\displaystyle \lim_{x\rightarrow x_0} f(x)=f(x_0)$,not just your teacher have said!

Also note that $\displaystyle \cos ,\sin\circ\cos$are both continuous at 0

Quote:

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Originally Posted by ynj

$\displaystyle \lim_{x\rightarrow 0}\frac{\sin (\cos x)}{\sec x}=\lim_{x\rightarrow 0}\sin (\cos x)\cos x=\lim_{x\rightarrow 0}\sin (\cos x)\lim_{x\rightarrow 0}\cos x$
$\displaystyle =\sin (\cos \lim_{x\rightarrow 0}x)\cos 0=\sin (\cos 0)\cos 0=\sin 1$
Note that there is also another theorem:
if $\displaystyle f(x)$is continuous at $\displaystyle x_0$, then
$\displaystyle \lim_{x\rightarrow x_0} f(x)=f(x_0)$,not just your teacher have said!

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Thank you so much (Cool)

Your teacher's point was that you had not yet learned that theorem and so shouldn't use it until you had. That is what was "theoretically" bad. (No, "theromatically", as good as it sounds, is NOT a word!)