# find the limit

• Aug 31st 2009, 09:01 PM
yoman360
find the limit
$\lim_{x\to0} \frac{sin(cos x)}{sec x}$

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)
• Aug 31st 2009, 09:05 PM
Chris L T521
Quote:

Originally Posted by yoman360
$\lim_{x\to0} \frac{sin(cos x)}{sec x}$

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

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

$\lim_{x\to0}\frac{\sin\left(\cos x\right)}{\sec x}=\frac{\sin\left(\cos 0\right)}{\sec 0}=\sin\left(1\right)$
• Aug 31st 2009, 09:13 PM
yoman360
Quote:

Originally Posted by Chris L T521
What's wrong with using direct substitution with trig functions?!?!

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

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.
• Aug 31st 2009, 09:24 PM
ynj
Quote:

Originally Posted by yoman360
$\lim_{x\to0} \frac{sin(cos x)}{sec x}$

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)

$\lim_{x\rightarrow 0}\frac{\sin (\cos x)}{\sec x}=\lim_{x\rightarrow 0}\sin (\cos x)\cos x=\lim_{x\rightarrow 0}\sin (\cos x)\lim_{x\rightarrow 0}\cos x$
$=\sin (\cos \lim_{x\rightarrow 0}x)\cos 0=\sin (\cos 0)\cos 0=\sin 1$
Note that there is also another theorem:
if $f(x)$is continuous at $x_0$, then
$\lim_{x\rightarrow x_0} f(x)=f(x_0)$,not just your teacher have said!
• Aug 31st 2009, 09:46 PM
ynj
Also note that $\cos ,\sin\circ\cos$are both continuous at 0
• Aug 31st 2009, 09:49 PM
yoman360
Quote:

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

Thank you so much (Cool)
• Sep 1st 2009, 07:46 AM
HallsofIvy
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!)