How Can You Prove Standard Limits Without Using L'Hopital's Rule?

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Discussion Overview

The discussion revolves around proving standard limits without using L'Hopital's rule, specifically the limits of log functions as x approaches 0 and infinity. The scope includes mathematical reasoning and homework-related inquiries.

Discussion Character

  • Exploratory, Homework-related, Mathematical reasoning

Main Points Raised

  • One participant seeks alternative methods to prove the limits: limit as x approaches 0 of log(1 + x)/x = 1 and limit as x approaches infinity of log(1 + x)/x = 0.
  • Another participant suggests using a series expansion as a potential method for proving the limits.
  • A different participant proposes using the squeeze theorem, providing a detailed approach for the first limit, demonstrating that log(1 + x) is bounded by x and x - (x^2)/2 as x approaches 0.
  • One participant expresses difficulty in finding sequences for the second limit that satisfy the conditions of the squeeze theorem.
  • Another participant attempts to use the limit definition involving e to show that the limit as x approaches 0 of log(1 + x)/x equals 1.
  • There is a suggestion that the function sqrt(x) could be used, but it is noted that it does not converge to 0, which raises further questions about finding appropriate bounding sequences.

Areas of Agreement / Disagreement

Participants generally agree on the need for alternative methods to prove the limits without L'Hopital's rule. However, there is no consensus on the approach for the second limit, as participants express differing levels of success and understanding in applying the squeeze theorem.

Contextual Notes

Some participants mention the challenge of finding appropriate sequences for the squeeze theorem, indicating potential limitations in their current understanding or approach. The discussion reflects varying levels of familiarity with mathematical techniques relevant to limits.

TheAvenger1
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I have a few questions for my homework assignments for solving limits, but in order to do those questions I have to use a few standard limits that we haven't been taught, which means I'll have to prove them. I know these can be done using L'Hopital's rule, but we haven't covered that yet so I was wondering whether there's some other way to prove these limits:

limit x -> 0 log (1 + x)/x = 1

and

limit x -> infinity log (1+x)/x = 0

Thanks in advance!
 
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What about using a series expansion , is it allowed ?
 
TheAvenger said:
I have a few questions for my homework assignments for solving limits, but in order to do those questions I have to use a few standard limits that we haven't been taught, which means I'll have to prove them. I know these can be done using L'Hopital's rule, but we haven't covered that yet so I was wondering whether there's some other way to prove these limits:

limit x -> 0 log (1 + x)/x = 1

and

limit x -> infinity log (1+x)/x = 0

Thanks in advance!

Hi Avenger! :)

Sure.
If you can find a sequence above and another sequence below your sequence, you can "squeeze" you sequence.
For your first limit, we have:

$\qquad x - \dfrac {x^2} 2 \le \log(1+x) \le x \qquad$

If x>0 we get:

$\qquad\dfrac{x - \dfrac {x^2} 2}{x} \le \dfrac{\log(1+x)}{x} \le \dfrac{x}{x} \qquad$

$\qquad1 - \dfrac {x} {2} \le \dfrac{\log(1+x)}{x} \le 1 \qquad$

If x approaches zero from above these expressions will approach 1, so

$\qquad\displaystyle\lim_{x \downarrow 0} \dfrac{\log(1+x)}{x} = 1$

Similarly you can prove that

$\qquad\displaystyle\lim_{x \uparrow 0} \dfrac{\log(1+x)}{x} = 1$

Therefore

$\qquad\displaystyle\lim_{x \to 0} \dfrac{\log(1+x)}{x} = 1$

Can you think of a similar way to do the second limit?
 
Ah of course, the sqeeze rule! I should have thought of that! I find it really difficult to come up with two sequences required to use the rule, but I'll try doing the second part myself before asking for assistance.

Thank you for the help!
 
Let $\displaystyle \ell = \lim_{x \to 0} \frac{\log(x+1)}{x}$ then $\displaystyle e^\ell = \lim_{x \to 0} (1+x)^{\frac{1}{x}}$. Let $x \mapsto \frac{1}{x}$ then $\displaystyle e^\ell = \lim_{x \to \infty}\left(1+\frac{1}{x}\right)^x = e.$ Thus $\displaystyle \ell = \log(e) = 1$.
 
I've tried thinking of two sequences to use to prove the second question, but I just can't come up with something that's always greater than log(1+x) which converges to 0. It should be simple enough but I'm hitting a brick wall...
 
TheAvenger said:
I've tried thinking of two sequences to use to prove the second question, but I just can't come up with something that's always greater than log(1+x) which converges to 0. It should be simple enough but I'm hitting a brick wall...

How about $\sqrt x$?

It does not converge to 0, but it is not supposed to.
Neither does $\log(1+x)$.
It is only supposed to increase slower than $x$, but faster than $\log(1+x)$.
 
TheAvenger said:
but we haven't covered that yet so I was wondering whether there's some other way to prove these

Perhaps, the following will be useful for you in the future: $$\lim_{x\to 0}\frac{\log (1+x)}{x}=\lim_{x\to 0}\frac{x+o(x)}{x}=\lim_{x\to 0}\left(1+\frac{o(x)}{x}\right)=1+0=1$$
 

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