Some special consequences from the limits of e: need a little help

[maxpayne_lhp]

Well, hello!
I have difficuties solving a problem.
Well, the first 3 ones are done successfully:
a. Calculate out the lim (x -> inf) of (1+2/x)^x
But assuming that 2/x = 2/t, I got the result to be e^2
b. Do the same thing with the lim (x ->0) of (1+3x)^(1/x)
Well, yeah "do the same thing", I assumed that t = 3x; I got the result of e^3
c.lim (x->0) of [(e^3x)-1]/x . Well, t = 3x this time. I got it as 3
But the third: d
lim (x->0) of { ln(1+sinx)/x }
Well, I tried assuming that t = sinx , but I got stuck in solving this, so far. Please give me a suggestion! Thanks!

 

[dextercioby]

Did u study derivatives and L'Hôspital's law??

Daniel.

P.S.The first 3 are perfect.

 

[dextercioby]

There are 2 approximations which u can use:

For "x" very,very small
\sin x\sim x
e^{x}\sim 1+x

Use these 2 approximations for good purpose...

Daniel.

 

[maxpayne_lhp]

No I haven't learned about it yet. Any other way?

 

[dextercioby]

Use the approximations.The first one can be proven via intuitive geometrical resoning and the second,well,it's a bit difficult...

Daniel.

P.S.The final result is "+1".

 

[HallsofIvy]

Well, hello!
I have difficuties solving a problem.
Well, the first 3 ones are done successfully:
a. Calculate out the lim (x -> inf) of (1+2/x)^x
But assuming that 2/x = 2/t, I got the result to be e^2

Changing x to t (as in 2/x= 2/t) doesn't do anything! If you have learned that lim(t->infinity) (1+ 1/t)^t= 2 then what you can do is let t= x/2. I'll bet that's what you actually did. Of course, then x= 2t and, clearly, as x-> inf, t-> inf. The limit becomes
lim(t->inf) (1+1/t)^2t= lim(t->inf) ((1+ 1/t)[sup t[/sup])² which, since x² is continuous, (is lim(t->inf) (1+1/t)^t)²= e².

b. Do the same thing with the lim (x ->0) of (1+3x)^(1/x)
Well, yeah "do the same thing", I assumed that t = 3x; I got the result of e^3

I assume that "(1+3x)" was really (1+ 3/x). If so, yes, that's correct.

c.lim (x->0) of [(e^3x)-1]/x . Well, t = 3x this time. I got it as 3

Is that e²x or e^3x? I assume the latter. Taking t= 3x, x= t/3 so this become lim(t->0)3(e^t-1)/t. (e^t-1)/t goes to 1 because it is the derivative of e^x at x= 0?

But the third: d
lim (x->0) of { ln(1+sinx)/x }
Well, I tried assuming that t = sinx , but I got stuck in solving this, so far. Please give me a suggestion! Thanks!

How about using the fact that ln(1+ sin x)/x= ln{(1+ sin x)^1/x}. Does that help? If we let y= ln{(1+sinx)^1/x}, then
e^y= (1+ sin x)^1/x. Can you find the limit of that?

 

[maxpayne_lhp]

Ok, I got it! I was so stupid!
Thanks, Mr. Hallsofivy, thanks Mr. Dextercoby

 

[maxpayne_lhp]

Sorry, misters, but I got stuck while finding the lim of e^y.

 

[dextercioby]

Halls said that
\lim_{x\rigtarrow 0} \frac{1}{x}\ln(1+\sin x)=\lim_{x\rightarrow 0} [\ln(1+\sin x)]^{\frac{1}{x}}

Denote the limit by "L" and exponentiate the previous relation.That's what Halls said.
e^{L}=\lim_{x\rightarrow 0}(1+\sin x)^{\frac{1}{x}}

And now use the trick:
\lim_{x\rightarrow 0} (1+x)^{\frac{1}{x}}=\lim_{\frac{1}{x}\rightarrow +\infty}(1+\frac{1}{\frac{1}{x}})^{\frac{1}{x}}=e

and the fact that
\lim_{x\rightarrow 0}\frac{\sin x}{x}=1

to find your answer.

Daniel.

 

[maxpayne_lhp]

Why e^{L}=\lim_{x\rightarrow 0}(1+\sin x)^{\frac{1}{x}} ?

 

[dextercioby]

Well,since:
L=\lim_{x\rightarrow 0} \ln(1+\sin x)^{\frac{1}{x}}

,what do you get by exponentiating the equation?

Daniel.

 

[maxpayne_lhp]

Well, I got it as lim (x->0) of [(1+sinx)^(1/x)]

 

[dextercioby]

Okay now use the trick with the reversing fractions and the of sine ratio.

Daniel.

 

[maxpayne_lhp]

Okay, thanks!
PS: Thanks for telling me about LaTex! It's cool! I'm learning it!

 

[maxpayne_lhp]

Hmmm... So far, I've done it. But I think tis way is better:

\lim_{x\rightarrow0} \frac{ln(1+sinx)}{x}

=\lim_{x\rightarrow0} \frac{ln(1+sinx)}{sinx} . \frac{sinx}{x}

=1.1 = 1

Is it right?

 

[dextercioby]

Nope.Can u prove that the first limit is "+1"??I frankly doubt it.

BTW,\sin x
The functions should not be italic...


Daniel.

 

[maverick280857]

If you have two functions which are both tending to zero as x tends to zero then they can be both replaced by equivalent infinitesimals. The approximation, \sin x = x near zero, would be useful here (though not without justification) if you do not know that \frac{ln(1+x)}{x} tends to 1 as x tends to zero or you are precluded from using the first form L'Hopital's Theorem.

 

[maxpayne_lhp]

Uhmm... alright!
Just imagine that sinx = \alpha
Then the thing above is equal to:

\lim_{x\rightarrow0}\frac{ln(1+\alpha)}{\alpha} (*)

Then, this limit is a consequence limit of e:

\lim_{x\rightarrow\infty}(1+\frac{1}{x})^x = e

So. (*) =1

 

[maxpayne_lhp]

Hmm... yeah, the basic consequence
\lim_{x\rightarrow0}\frac{ln(1+u)}{u}=1