# Evaluate a limit

• Aug 6th 2013, 01:07 PM
Evaluate a limit
Hey everyone,

Got this problem from a math packet. No idea how to evaluate this limit. I assume L'hopital's rule is involved.

Attachment 28945

No matter how hard I try, I am unable to manipulate this limit in such a way as to evaluate it.
• Aug 6th 2013, 01:11 PM
MarkFL
Re: Evaluate a limit
Have you tried taking the natural log, so that you get the form 0/0?
• Aug 6th 2013, 03:13 PM
HallsofIvy
Re: Evaluate a limit
First obvious step is to factor out $\frac{1}{2}$. I would then factor out $a^x$ so this becomes $\frac{a}{2}(1+ (b/a)^x)^{1/x}$.

Now, do you know the limit definition of "e"?
• Aug 6th 2013, 03:18 PM
MarkFL
Re: Evaluate a limit
Quote:

Originally Posted by HallsofIvy
First obvious step is to factor out $\frac{1}{2}$. I would then factor out $a^x$ so this becomes $\frac{a}{2}(1+ (b/a)^x)^{1/x}$.

Now, do you know the limit definition of "e"?

I don't think you can factor out $\frac{1}{2}$.
• Aug 6th 2013, 05:41 PM
Plato
Re: Evaluate a limit
Quote:

Originally Posted by HallsofIvy
First obvious step is to factor out $\frac{1}{2}$. I would then factor out $a^x$ so this becomes $\frac{a}{2}(1+ (b/a)^x)^{1/x}$

Quote:

Originally Posted by MarkFL
I don't think you can factor out $\frac{1}{2}$.

That is correct. It cannot be factored in that way.

[URL="[QUOTE=HallsofIvy;794322]First obvious step is to factor out $\frac{1}{2}$. I would then factor out $a^x$ so this becomes $\frac{a}{2}(1+ (b/a)^x)^{1/x}$. Now, do you know the limit definition of "e"?[/QUOTE]"

http://www.wolframalpha.com/input/?i...9%2F%28x%29%29
• Aug 21st 2013, 06:28 AM
x3bnm
Re: Evaluate a limit
Quote:

Hey everyone,

Got this problem from a math packet. No idea how to evaluate this limit. I assume L'hopital's rule is involved.

Attachment 28945

No matter how hard I try, I am unable to manipulate this limit in such a way as to evaluate it.

\begin{align*}&\lim_{x\to 0}\left(\frac{a^x + b^x}{2}\right)^{\frac{1}{x}}\\=& \lim_{x\to 0}e^{\left(\ln\left(\frac{a^x + b^x}{2}\right)^{\frac{1}{x}}\right)}\\=& \lim_{x\to 0} e^{\left(\frac{1}{x}\ln{\left(a^x + b^x\right)} - \frac{1}{x}\ln{2}\right)}\\ & \\=& \lim_{x\to 0} \frac{ \left( e^{\frac{1}{x}} \right) \left(e^{\ln{\left(a^x + b^x \right)}}\right)}{\left( e^{\frac{1}{x}}\right) \left(e^{\ln{2}}\right)}\\& \\=& \lim_{x\to 0} \frac{\left(a^x + b^x\right)}{2}\\=& \frac{(1+1)}{2} \\=& 1........\text{[Answer]}\end{align*}

Hope it helps.
• Aug 21st 2013, 06:49 AM
Plato
Re: Evaluate a limit
Quote:

Originally Posted by x3bnm
\begin{align*}&\lim_{x\to 0}\left(\frac{a^x + b^x}{2}\right)^{\frac{1}{x}}\\=& \lim_{x\to 0}e^{\left(\ln\left(\frac{a^x + b^x}{2}\right)^{\frac{1}{x}}\right)}\\=& \lim_{x\to 0} e^{\left(\frac{1}{x}\ln{\left(a^x + b^x\right)} - \frac{1}{x}\ln{2}\right)}\\ & \\=& \lim_{x\to 0}\color{red} \frac{ \left( e^{\frac{1}{x}} \right) \left(e^{\ln{\left(a^x + b^x \right)}}\right)}{\left( e^{\frac{1}{x}}\right) \left(e^{\ln{2}}\right)}\\& \end{align*}

@x3bnm;795132
$\left( e^{\frac{1}{x}} \right) \left(e^{\ln{\left(a^x + b^x \right)}}\right)\ne e^{\left(\frac{1}{x}\ln{\left(a^x + b^x\right)} \right)}$

We add exponents not multiply them.
• Aug 21st 2013, 10:03 AM
BobP
Re: Evaluate a limit
This is a bit rough and ready, but I think that the result is okay.

Suppose that $b>a,$ then

$\left(\frac{a^{x}+b^{x}}{2}\right)^{1/x}=b\left(\frac{(a/b)^{x}+1}{2}\right)^{1/x}\rightarrow b,$ as $x \rightarrow \infty,$
(since $(a/b)^{x}\rightarrow 0,$ and the resulting $(1/2)^{1/x}\rightarrow 1).$

Notice btw that if $a=b$ then

$\left(\frac{a^{x}+b^{x}}{2}\right)^{1/x}=a,$ (or $b$).
• Aug 21st 2013, 01:17 PM
Plato
Re: Evaluate a limit
Quote:

Originally Posted by BobP
This is a bit rough and ready, but I think that the result is okay.

Suppose that $b>a,$ then

$\left(\frac{a^{x}+b^{x}}{2}\right)^{1/x}=b\left(\frac{(a/b)^{x}+1}{2}\right)^{1/x}\rightarrow b,$ as $x \rightarrow \infty,$
(since $(a/b)^{x}\rightarrow 0,$ and the resulting $(1/2)^{1/x}\rightarrow 1).$

Notice btw that if $a=b$ then

$\left(\frac{a^{x}+b^{x}}{2}\right)^{1/x}=a,$ (or $b$).

NOTE $(a/b)^{x}\not\rightarrow 0,$ you have $(a/b)^{x}\rightarrow 1,$

On the wolfram page l'Hospitial's rule gives $\exp\left(\frac{\ln(ab)}{2}\right)$
• Aug 21st 2013, 02:27 PM
x3bnm
Re: Evaluate a limit
Quote:

Hey everyone,

Got this problem from a math packet. No idea how to evaluate this limit. I assume L'hopital's rule is involved.

Attachment 28945

No matter how hard I try, I am unable to manipulate this limit in such a way as to evaluate it.

\begin{align*}&\lim_{x\to 0} \left(\frac{(a^x + b^x)}{2}\right)^{\frac{1}{x}}\\=& \lim_{x\to 0} e^{\ln{\left(\frac{(a^x + b^x)}{2}\right)^{\frac{1}{x}}}}\\=& \lim_{x\to 0} e^{\frac{\ln{\left(a^x + b^x\right)}-\ln(2)}{x}}\end{align*}

Now using I'Hopital's rule and property of limit this last line becomes:

\begin{align*}&e^{\lim_{x\to 0} \frac{a^x \ln{(a)} + b^x \ln{(b)}}{a^x + b^x}}\\=& e^{\lim_{x\to 0} \frac{a^0 \ln{(a)} + b^0 \ln{(b)}}{a^0 + b^0}}\\=& e^{\left(\frac{\ln{(ab)}}{2}\right)}\\=& ab - e^2\text{.......[Answer]}\end{align*}
• Aug 21st 2013, 02:35 PM
Re: Evaluate a limit
thanks x3bnm. The final step actually tursn the answer to sqrt(a)*sqrt(b) rather than ab-e^2, the process helped a lot.
• Aug 21st 2013, 02:43 PM
Plato
Re: Evaluate a limit
Quote:

Originally Posted by x3bnm
\begin{align*}&\lim_{x\to 0} \left(\frac{(a^x + b^x)}{2}\right)^{\frac{1}{x}}\\=& \lim_{x\to 0} e^{\ln{\left(\frac{(a^x + b^x)}{2}\right)^{\frac{1}{x}}}}\\=& \lim_{x\to 0} e^{\frac{\ln{\left(a^x + b^x\right)}-\ln(2)}{x}}\end{align*}

Now using I'Hopital's rule and property of limit this last line becomes:

\color{red}\begin{align*}&e^{\lim_{x\to 0} \frac{a^x \ln{(a)} + b^x \ln{(b)}}{a^x + b^x}}\\=& e^{\lim_{x\to 0} \frac{a^0 \ln{(a)} + b^0 \ln{(b)}}{a^0 + b^0}}\\=& e^{\left(\frac{\ln{(ab)}}{2}\right)}\\=& ab - e^2\text{.......[Answer]}\end{align*} No indeed!

$\exp{\left(\frac{\ln{(ab)}}{2}\right)}=\exp\left( \ln{(\sqrt{ab}) \right)}=\sqrt{ab}$

@x3bnm: Please make sure that you understand the process before posting.
Did you look at either of the Wolfram webpages links that I posted in replies #5 & #9?
If not, why not?
• Aug 22nd 2013, 12:11 AM
BobP
Re: Evaluate a limit
Hi Plato

Regarding your comment to my post #8, your post #9, somewhat sheepishly I have to admit that, (have just noticed that), I was calculating the wrong limit, $x\rightarrow \infty$ rather than $x\rightarrow 0.$

Apologies to all.