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

1. Jan 21, 2005

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!

2. Jan 21, 2005

dextercioby

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

Daniel.

P.S.The first 3 are perfect.

3. Jan 21, 2005

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.

4. Jan 21, 2005

maxpayne_lhp

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

5. Jan 21, 2005

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".

6. Jan 21, 2005

HallsofIvy

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.
I assume that "(1+3x)" was really (1+ 3/x). If so, yes, that's correct.
Is that e2x or e3x? I assume the latter. Taking t= 3x, x= t/3 so this become lim(t->0)3(et-1)/t. (et-1)/t goes to 1 because it is the derivative of ex at x= 0?
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
ey= (1+ sin x)1/x. Can you find the limit of that?

7. Jan 21, 2005

maxpayne_lhp

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

Last edited: Jan 21, 2005
8. Jan 21, 2005

maxpayne_lhp

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

9. Jan 21, 2005

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
$$\lim_{x\rightarrow 0}\frac{\sin x}{x}=1$$

Daniel.

10. Jan 21, 2005

maxpayne_lhp

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

11. Jan 21, 2005

dextercioby

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

,what do you get by exponentiating the equation?

Daniel.

12. Jan 21, 2005

maxpayne_lhp

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

13. Jan 21, 2005

dextercioby

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

Daniel.

14. Jan 21, 2005

maxpayne_lhp

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

15. Jan 21, 2005

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?

Last edited: Jan 21, 2005
16. Jan 21, 2005

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.

17. Jan 22, 2005

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.

18. Jan 22, 2005

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

19. Jan 22, 2005

maxpayne_lhp

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