Limits?

[LeighCapehart]

I've been able to get by on several problems just by looking at the graphs, however, this problem is giving me alot of trouble. Could someone please explain to me how to do this problem, and an easier way to solve it than just guessing based on graphs?

Thanks

---Leigh

 

[lookagain]

\(\displaystyle Let \ y \ = \ the \ expression \ in \ the \ quote \ box.\)


Take the natural logs of each side:

\(\displaystyle ln(y) = ln\bigg[\lim_{x \to 0}(1 + x)^{\frac{1}{x}}\bigg]\)


\(\displaystyle ln(y) = \lim_{x \to 0}\bigg \{ln \bigg[(1 + x)^{\frac{1}{x}} \bigg]\bigg\}\)


Bring the exponent (1/x) down as a multiplier:

\(\displaystyle ln(y) = \lim_{x \to 0} \bigg[\bigg(\frac{1}{x} \bigg)ln(1 + x)\bigg]\)

Rewrite:

\(\displaystyle ln(y) = \lim_{x \to 0}\frac{ln(1 + x)}{x}\)

One of the possibilities is to use L'Hopital's Rule, as the expression
on the right-hand side is one of the indeterminate forms:

\(\displaystyle ln(y) = \lim_{x \to 0}\frac{1/(1 + x)}{1}\)


\(\displaystyle ln(y) = \frac{1/(1 + 0)}{1} = \frac{1}{1}\)


\(\displaystyle ln(y) = 1\)


\(\displaystyle Raise \ the \ constant \ e \ to \ each \ side:\)


\(\displaystyle e^{ln(y)} \ = \ e^1\)


\(\displaystyle y \ = \ e\)

\(\displaystyle So \ \ the \ \lim_{x \to 0}{(1 + x)^{\frac{1}{x}} \ = \ \boxed{e}\)

 

[galactus]

If I may, here is another way to approach the problem without L'Hopital.

As lookagain showed, L'Hopital is a handy tool. Thus, many difficult limits can be evaluated using it.

Consider the differentiability of ln(x) at x=1.

\(\displaystyle \left \frac{d}{dx}[ln(x)]\right|_{x=1}=\left \frac{1}{x}\right|_{x=1}=1\)

Using the defintion of a derivative:

\(\displaystyle 1=\lim_{h\to 0}\frac{ln(1+h)-ln(1)}{h}=\lim_{h\to 0}\frac{ln(1+h)}{h}=\lim_{h\to 0} ln(1+h)^{\frac{1}{h}\)

e to both sides:

\(\displaystyle e^{1}=\large e^{\lim_{h\to 0} ln(1+h)^{\frac{1}{h}}}\)

Since e^x is continuous, we can write it as:

\(\displaystyle e=\lim_{h\to 0} e^{ln(1+h)^{\frac{1}{h}}}=\underbrace{\lim_{h\to 0}(1+h)^{\frac{1}{h}}}_{\text{what we started with}}\)

 

[lookagain]

---Leigh

Another way:

Look at the \(\displaystyle (1 + x)^{\frac{1}{x}}\) part.

Using the binomial formula, expand it as:

\(\displaystyle \frac{1}{0!}(1) + \frac{1}{1!} \bigg[\frac{1}{x}\bigg({x}\bigg) \bigg] + \frac{1}{2!}\bigg[\frac{1}{x}\bigg(\frac{1}{x} - 1\bigg)(x^2)\bigg] + ... \ =\)


\(\displaystyle 1 + \frac{1}{1!} + \frac{1}{2!}\bigg(1 - x \bigg) + \frac{1}{3!}\bigg(1 \ added \ to \ x \ terms \ of \ degree \ 1\ or\ higher \bigg) + ...\)

\(\displaystyle + \frac{1}{n!}\bigg(1 \ added\ to \ x \ terms \ of \ degree \ 1 \ or \ higher \bigg)** \ \ + ...\)


\(\displaystyle Then \ the \ limit \ of \ this \ expression \ from \ the \ last \ line \ as \ x \rightarrow \ 0 \ is:\)

\(\displaystyle \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + ... \ =\)


\(\displaystyle \boxed{e}\)



\(\displaystyle ** This \ is \ the \ general \ term.\)