Limits of functions (Real Analysis)

[trickslapper]

Evaluate the Limit of the following:

tan(?x)/x+2, as x approaches negative 2.

(3* (³?x) - 2?x)[sup:214m3ffh]ln(x)[/sup:214m3ffh], as x approaches 1

Our professor told us to use the following results that we proved in class:

1. If v(x) and u(x) have limits, call them A and B (respectively), Then the Limit of u(x)[sup:214m3ffh]v(x)[/sup:214m3ffh] as x approaches x[sub:214m3ffh]0[/sub:214m3ffh] = Lim u(x)[sup:214m3ffh]Lim v(x)[/sup:214m3ffh], as x approaches x[sub:214m3ffh]0[/sub:214m3ffh] for both limits.
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2. IF we have something of the form u(x)[sup:214m3ffh]v(x)[/sup:214m3ffh], we can rewrite it as: e[sup:214m3ffh]xln(v)[/sup:214m3ffh]


*My result number 2 may not be totally right, i missed class that day and i had to copy from someones notes, so i'm not sure that its exactly what my professor wrote.

anyways can someone push me in the right direction? thanks!

 

[galactus]

\(\displaystyle \lim_{x\to -2}\frac{tan({\pi}x)}{x+2}\)

Is this what you meant?. You have \(\displaystyle \frac{tan({\pi}x)}{x}+2\) written. Grouping symbols are important. You are in calc, so..........

Are you allowed to use L'Hopital's Rule?. If not, we can find another approach. But if using L'Hopital, then the derivative of \(\displaystyle tan({\pi}x)\) is

\(\displaystyle {\pi}sec^{2}({\pi}x)\). The derivative of x+2 is simply 1. So, take the limit of \(\displaystyle {\pi}sec^{2}({\pi}x)\) by simply plugging in x=-2 and see what you get.

\(\displaystyle \lim_{x\to 1}(3\sqrt[3]{x} - 2\sqrt{x})^{ln(x)}\)

Your professor meant that if you have u(x) and v(x) that have limits, then \(\displaystyle \lim_{x\to 1}(3\sqrt[3]{x}-2\sqrt{x})^{ln(x)}=\lim_{x\to 1}(3\sqrt[3]{x}-2\sqrt{x})^{\lim_{x\to 1}ln(x)}\)

Using the law your professor gave shows that the limit on the exponent becomes ln(1)=0.

So, we have \(\displaystyle (1)^{0}=1\)
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2. IF we have something of the form u(x)[sup:27zivtrh]v(x)[/sup:27zivtrh], we can rewrite it as: e[sup:27zivtrh]xln(v)[/sup:27zivtrh]

Note that \(\displaystyle e^{xln(v)}=e^{ln(v^{x})}=v^{x}\)


*My result number 2 may not be totally right, i missed class that day and i had to copy from someones notes, so i'm not sure that its exactly what my professor wrote.

anyways can someone push me in the right direction? thanks![/quote]

 

[trickslapper]

I'm sorry i meant that the (x+2) term was in the denominator, and our professor says we don't know anything about derivatives therefore we aren't allowed to use l'hopitals, only the two rules he gave us.

thanks for the help so far!

 

[trickslapper]

Also, i just spoke with my professor and none of the cases he showed us apply to the first problem. It's just a find the limit without using l'hopitals rule

 

[daon]

\(\displaystyle \frac{\tan (\pi x)}{x+2} = \pi \frac{\tan (\pi (x+2))}{\pi (x+2)}\) ...why?

Then think a little bit and you'll be able to use \(\displaystyle \lim_{h \to 0} \frac{\sin(h)}{h} = 1\)

 

[trickslapper]

Okay i understand the first part of your response because: tan( ?)=tan( ?+n ?) for even n..I think.

but what gave you the idea to multiply by ? on top and bottom?
EDIT: Nevermind i see why you multiply by ?, so you can use the sin(x)/x =1 as x approaches 0 rule.

I ended up getting ? as my answer because i eventually got to: ? * sin(?h)/?h * (1/cos(?h)) as h approaches 0, which turns into ?*1*1=?. Is this right?

 

[daon]

Okay i understand the first part of your response because: tan( ?)=tan( ?+n ?) for even n..I think.

but what gave you the idea to multiply by ? on top and bottom?
EDIT: Nevermind i see why you multiply by ?, so you can use the sin(x)/x =1 as x approaches 0 rule.

I ended up getting ? as my answer because i eventually got to: ? * sin(?h)/?h * (1/cos(?h)) as h approaches 0, which turns into ?*1*1=?. Is this right?

Yes, but take note that tan has period Pi, not 2Pi, so the same idea would work if the denominator had been x+1.

 

[trickslapper]

ok cool, thanks for the help!