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

[TheAvenger1]

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!

 

[alyafey22]

What about using a series expansion , is it allowed ?

 

[I like Serena]

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?

 

[TheAvenger1]

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!

 

[Moriarty1]

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 $\displaystyle \ell = \log(e) = 1$.

 

[TheAvenger1]

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...

 

[I like Serena]

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 $\log(1+x)$.

 

[Fernando Revilla]

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: $$\lim_{x\to 0}\frac{\log (1+x)}{x}=\lim_{x\to 0}\frac{x+o(x)}{x}=\lim_{x\to 0}\left(1+\frac{o(x)}{x}\right)=1+0=1$$