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

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In summary, Daniel says that the first three limits are all correct, but the last limit is not correct because he was stuck when trying to find the limit of e^y. He provides two approximations which can be used to solve the problem. The first is when x is very small, and the second is when t is x/2. The final result is "+1".
  • #1
maxpayne_lhp
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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!
 
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  • #2
Did u study derivatives and L'Hôspital's law??

Daniel.

P.S.The first 3 are perfect.
 
  • #3
There are 2 approximations which u can use:

For "x" very,very small
[tex] \sin x\sim x [/tex]
[tex] e^{x}\sim 1+x [/tex]

Use these 2 approximations for good purpose...

Daniel.
 
  • #4
No I haven't learned about it yet. Any other way?
 
  • #5
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
maxpayne_lhp said:
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 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?
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
ey= (1+ sin x)1/x. Can you find the limit of that?
 
  • #7
Ok, I got it! I was so stupid!
Thanks, Mr. Hallsofivy, thanks Mr. Dextercoby
 
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  • #8
Sorry, misters, but I got stuck while finding the lim of e^y.
 
  • #9
Halls said that
[tex] \lim_{x\rigtarrow 0} \frac{1}{x}\ln(1+\sin x)=\lim_{x\rightarrow 0} [\ln(1+\sin x)]^{\frac{1}{x}} [/tex]

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

And now use the trick:
[tex] \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 [/tex]

and the fact that
[tex] \lim_{x\rightarrow 0}\frac{\sin x}{x}=1 [/tex]

to find your answer.

Daniel.
 
  • #10
Why [tex] e^{L}=\lim_{x\rightarrow 0}(1+\sin x)^{\frac{1}{x}} [/tex] ?
 
  • #11
Well,since:
[tex] L=\lim_{x\rightarrow 0} \ln(1+\sin x)^{\frac{1}{x}} [/tex]

,what do you get by exponentiating the equation?

Daniel.
 
  • #12
Well, I got it as lim (x->0) of [(1+sinx)^(1/x)]
 
  • #13
Okay now use the trick with the reversing fractions and the of sine ratio.

Daniel.
 
  • #14
Okay, thanks!
PS: Thanks for telling me about LaTex! It's cool! I'm learning it!
 
  • #15
Hmmm... So far, I've done it. But I think tis way is better:

[tex]\lim_{x\rightarrow0} \frac{ln(1+sinx)}{x}[/tex]

=[tex]\lim_{x\rightarrow0} \frac{ln(1+sinx)}{sinx} . \frac{sinx}{x}[/tex]

=[tex]1.1 = 1[/tex]

Is it right?
 
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  • #16
Nope.Can u prove that the first limit is "+1"??I frankly doubt it.

BTW,[tex] \sin x [/tex]
The functions should not be italic...


Daniel.
 
  • #17
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, [itex]\sin x = x[/itex] near zero, would be useful here (though not without justification) if you do not know that [itex]\frac{ln(1+x)}{x}[/itex] tends to 1 as x tends to zero or you are precluded from using the first form L'Hopital's Theorem.
 
  • #18
Uhmm... alright!
Just imagine that [tex]sinx = \alpha[/tex]
Then the thing above is equal to:

[tex]\lim_{x\rightarrow0}\frac{ln(1+\alpha)}{\alpha}[/tex] (*)

Then, this limit is a consequence limit of e:

[tex]\lim_{x\rightarrow\infty}(1+\frac{1}{x})^x = e[/tex]

So. (*) =1
 
  • #19
Hmm... yeah, the basic consequence
[tex]\lim_{x\rightarrow0}\frac{ln(1+u)}{u}=1[/tex]
 

1. What are the limits of e?

The limit of e, also known as Euler's number, is an irrational number approximately equal to 2.71828. It is the base of the natural logarithm and has many applications in mathematics, physics, and other sciences.

2. What are some special consequences of the limits of e?

One special consequence of the limits of e is its relationship to compound interest and continuous growth. Another consequence is its use in exponential functions and their derivatives.

3. How is the limit of e calculated?

The limit of e can be calculated using the infinite series expansion: e = 1 + 1/1! + 1/2! + 1/3! + ...

4. What is the significance of e in calculus?

In calculus, e is important in the study of continuous growth and decay. It is also used in the definition of the natural logarithm and in solving differential equations.

5. Can the limit of e be exceeded?

No, the limit of e cannot be exceeded as it is an infinite value. However, it can be approximated to any desired degree of accuracy.

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