Example dealing w/ a Trig Limit

[azwethinkweiz]

lim of x-->0[sup:21xjrc59]+[/sup:21xjrc59] (1/x - 1/sin x)

I then wrote the expression as follows: (sin x - x / x sin x)

and then decided to use L'Hospital's Rule as follows:

lim of x-->0[sup:21xjrc59]+[/sup:21xjrc59] (cos x - 1) / (xcosx + sinx)

and then I differentiated the numerator and denominator again which results in:

lim of x-->0[sup:21xjrc59]+[/sup:21xjrc59] (-sinx ) / (cosx - xsinx + cosx)

which results in the limit being 0.

But I'm not sure if this is correct :?

 

[BigGlenntheHeavy]

It is correct.

 

[azwethinkweiz]

Ah alright, thanks so much.

What if an expression is raised to the power of a trig function? For example,

lim as x->0 for ( 2/3 + e[sup:3c2uv7jv]-x[/sup:3c2uv7jv]/3)[sup:3c2uv7jv]cot x[/sup:3c2uv7jv]

[As in, the entire function raised to the power of cot x....]

I decided to approach this from a logarithmic point of view... by setting y = ( (2+e[sup:3c2uv7jv]-x[/sup:3c2uv7jv])/3)[sup:3c2uv7jv]cot x[/sup:3c2uv7jv]

which then leads to:

ln y = cot x ln (2+e[sup:3c2uv7jv]-x[/sup:3c2uv7jv])/3
ln y = cot x [ln 2 + ln e[sup:3c2uv7jv]-x[/sup:3c2uv7jv] - ln 3]

But what next? How do you bring this back to the original problem (provided I am on the right track)?

 

[daon]

For that limit, you can rewrite it as:

\(\displaystyle lny=\frac{cosx \cdot ln(\frac{2+e^{-x}}{3})}{sin(x)} \to \frac{ 0}{0}\)

So you may apply L'Hopital's Rule, but it won't be pretty.

After that, and your limit now exists, apply e. If you need to, try it again. I didn't bother to see if L'Hopital's will continue to work. Good luck

 

[azwethinkweiz]

Ah alright But if I do apply L'Hopital's Rule.. do I differentiate both sides? (as in the ln y and the other side..) Wouldnt that bring in implicit differentiation though?? That's the part where I'm confused.

 

[galactus]

You could apply L'Hopital twice. No, there is no implicit differentiation.

Rewrite as \(\displaystyle \lim_{x\to 0^{+}}\frac{sin(x)-x}{xsin(x)}\)

Apply L'Hopital:

\(\displaystyle \lim_{x\to 0^{+}}\frac{cos(x)-1}{xcos(x)+sin(x)}\)

Again:

\(\displaystyle \lim_{x\to 0^{+}}\frac{-sin(x)}{2cos(x)-xsin(x)}=0\)

Just as you done. That was good. Glenn said it was OK...it's OK.

 

[azwethinkweiz]

Yuup. Sorry, I got that, but thanks though.

I actually posted another problem regarding trig limits in the same discussion topic... (I'm new to this - are we supposed to do it in a different one? :? ) - but anyway... my second reply was to do with this problem:

What if an expression is raised to the power of a trig function? For example,

lim as x->0 for ( 2/3 + e-x/3)cot x

(I have explained what I tried in the reply above.)

 

[galactus]

That e-x/3 looks like \(\displaystyle e-\frac{x}{3}\). I assume you mean \(\displaystyle \frac{e^{-x}}{3}\)?. Please use proper grouping symbols. Otherwise, it may be misinterpreted.

\(\displaystyle \lim_{x\to 0}\left(\frac{2+e^{-x}}{3}\right)^{cot(x)}\)

Is that the correct problem before I go further?.

 

[azwethinkweiz]

Yeah, that's right.. Sorry about that, I'll familiarize myself with the proper grouping symbols.. but anyways, you got the expression right.

 

[galactus]

The site is kind of acting cranky this morning. If I did not respond, it is because it went down for some reason.

Anyway, rewrite as:

\(\displaystyle \lim_{x\to 0} \large e^{\left(cot(x)ln(\frac{2}{3}+\frac{1}{3}e^{-x})\right)}\)

\(\displaystyle \large e^{\left(\lim_{x\to 0}cot(x)ln(\frac{2}{3}+\frac{1}{3}e^{-x})\right)}\)

Now, use L'Hopital on the limit in the e power:

\(\displaystyle \large e^{\left(-lim_{x\to 0}\frac{e^{-x}cot^{2}(x)}{(2+e^{-x})(1+cot^{2}(x))}\right)}\)

After doing some algebra, we get it down to:

\(\displaystyle e^{\left(\frac{-1}{3}\lim_{x\to 0}\frac{cot^{2}(x)}{1+cot^{2}(x)}\cdot \lim_{x\to 0}e^{-x}\right)}\)

Now, use L'Hopital again and get:

\(\displaystyle e^{\left(\frac{-1}{3}\lim_{x\to 0}1\cdot \lim_{x\to 0}e^{-x}\right)}\)

\(\displaystyle e^{\left(\frac{-1}{3}\lim_{x\to 0} e^{-x}\right)}\)

Now, you can see the limit?. I know I skipped some algebra parts. See if you can work that out. Remember your identities.

Such as \(\displaystyle \frac{cot^{2}(x)}{1+cot^{2}(x)}=cos^{2}(x)\)

 

[azwethinkweiz]

That definitely helps, thanks so much!