find the limit

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August 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)
August 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?!?!

$\lim_{x\to0}\frac{\sin\left(\cos x\right)}{\sec x}=\frac{\sin\left(\cos 0\right)}{\sec 0}=\sin\left(1\right)$
August 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.
August 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!
August 31st 2009, 09:46 PM
ynj

Also note that $\cos ,\sin\circ\cos$ are both continuous at 0
August 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)
September 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!)