[SOLVED] Exponential Limits

**Hweengee** · October 10th 2008, 09:00 PM

Hi guys

I'm at the stage of my calculus course where we've just been introduced to the logarithm and exponential functions and I'm having some trouble evaluating limits in the form of exponential functions.

lim as x->0 (exp(2x)+2x)^1/x, so far i've gotten up to lim x->0 (1+2x/exp(2x))^1/x*e^2

lim as x->0 (((1+x+x^2)^1/x)/exp(1))^1/x

I know that the number e can be defined as the limit as x goes to 0 of (1+x)^1/x, but I'm at a complete loss when it comes to applying this to evaluating the given limits.

Any help would be appreciated.

Thanks.

**galactus** · October 11th 2008, 03:47 AM

Here is a quicky way, so to speak. We can do it without, but what the heck.

$\lim_{x\to 0}\left(e^{2x}+2x\right)^{\frac{1}{x}}$

Let t=2x:

$\lim_{t\to 0}\left(e^{t}+t\right)^{\frac{2}{t}}$

Rewrite:

$\displaystyle{^{2\lim_{t\to 0}\frac{ln(e^{t}+t)}{t}}}$

Now, we'll take the easy way out and use L'Hopital, giving:

$\displaystyle{e^{2\lim_{t\to 0}\frac{e^{t}+1}{e^{t}+t}}}$

Now, the limit can be seen by merely plugging in t=0.

We could also do it without the t substitution by rewriting it as:

$e^{\lim_{x\to 0}\frac{ln(e^{2x}+2x)}{x}}$

**Krizalid** · October 11th 2008, 07:02 AM

No need L'Hôpital, note that,

$\begin{aligned} \left( e^{2x}+2x \right)^{1/x}&=\exp \left\{ \frac{\ln \left( e^{2x}+2x \right)}{x} \right\} \\ & =\exp \left\{ \frac{2x+\ln \left( 1+\dfrac{2x}{e^{2x}} \right)}{x} \right\} \\ & =\exp \left\{ 2+\frac{\ln \left( 1+\dfrac{2x}{e^{2x}} \right)}{\dfrac{2x}{e^{2x}}}\cdot \frac{2}{e^{2x}} \right\}, \end{aligned}$

hence this last expression is $e^4$ as $x\to0$ and we're done.

**galactus** · October 11th 2008, 07:25 AM

No need L'Hôpital

I know K-meister. I was just lazy.

It's always nice to do them without if possible. I wonder why we dissuade ourselves from using the Hospital rule?.

It makes things easier. It's just a thing I suppose. I like to veer away from it, too, most of the time.

**Mathstud28** · October 11th 2008, 08:15 AM

$\left(e^{2x}+2x\right)^{\frac{1}{x}}=e^{\frac{\ln( e^{2x}+2x)}{x}}\sim{e^{\frac{\ln(1+2x+2x)}{x}}}=e^ {\frac{\ln(1+4x)}{x}}\sim{e^{\frac{4x}{x}}}=e^{4}$

**RedBarchetta** · October 11th 2008, 03:28 PM

Originally Posted by Mathstud28

$\left(e^{2x}+2x\right)^{\frac{1}{x}}=e^{\frac{\ln( e^{2x}+2x)}{x}}\sim{e^{\frac{\ln(1+2x+2x)}{x}}}=e^ {\frac{\ln(1+4x)}{x}}\sim{e^{\frac{4x}{x}}}=e^{4}$

How do you get from step 2 to step 3?

**Mathstud28** · October 11th 2008, 05:39 PM

Originally Posted by RedBarchetta

How do you get from step 2 to step 3?

Do you know about asymptotic equivalences?

**RedBarchetta** · October 11th 2008, 05:44 PM

Originally Posted by Mathstud28

Do you know about asymptotic equivalences?

No.......? I'm halfway through Calculus II.

**Hweengee** · October 11th 2008, 09:33 PM

Originally Posted by Krizalid

No need L'Hôpital, note that,

$\begin{aligned} exp \left\{ 2+\frac{\ln \left( 1+\dfrac{2x}{e^{2x}} \right)}{\dfrac{2x}{e^{2x}}}\cdot \frac{2}{e^{2x}} \right\}, \end{aligned}$

hence this last expression is $e^4$ as $x\to0$ and we're done.

sorry, i still dont get this last bit. how does the expression become exp(4) when x goes to 0? doesn't the denominator of the expression on the right become 0 as x goes to 0? so how does the expression on the right become 2 as x goes to 0?

**Moo** · October 12th 2008, 02:28 AM

Hello,

Originally Posted by Hweengee

sorry, i still dont get this last bit. how does the expression become exp(4) when x goes to 0? doesn't the denominator of the expression on the right become 0 as x goes to 0? so how does the expression on the right become 2 as x goes to 0?

$\exp(0)=e^0=1$

So $\frac{2}{e^{2x}} \stackrel{x \to 0}{\longrightarrow} \frac 21=2$

Moreover, $\dfrac{\ln \left(1+\tfrac{2x}{e^{2x}}\right)}{\tfrac{2x}{e^{2 x}}} \stackrel{x \to 0}{\longrightarrow} 1$ , because if you make a substitution $t=\frac{2x}{e^{2x}}$ , you'll have :

$\frac{\ln(1+t)}{t}$ , which tends to 1 as t tends to 0.

**Hweengee** · October 12th 2008, 02:47 AM

Originally Posted by Moo

$\frac{\ln(1+t)}{t}$ , which tends to 1 as t tends to 0.

Hi, is this bit by definition? since the number e is defined as the limit of (1+x)^1/x? and ln e is 1? Thanks for showing me how it works out, I was really lost before this.

**Moo** · October 12th 2008, 03:17 AM

Originally Posted by Hweengee

Hi, is this bit by definition? since the number e is defined as the limit of (1+x)^1/x? and ln e is 1? Thanks for showing me how it works out, I was really lost before this.

Hmmm here is my method =)

I'll use this definition : $f'(a)=\lim_{t \to 0} ~ \frac{f(a+t)-f(a)}{t}$

Now, let $f(x)=\ln(1+x)$
You can see that $f(0)=\ln(1+0)=\ln(1)=0$

So $\lim_{t \to 0} ~ \frac{\ln(1+t)}{t}=\lim_{t \to 0} ~ \frac{\ln(1+t)-0}{t}=\lim_{t \to 0} ~ \frac{\ln(1+t)-\ln(1)}{t}=\lim_{t \to 0} ~ \frac{f(t)-f(0)}{t}$

$=f'(0)=\left(\ln(1+t)\right)'(0)=\frac{1}{1+0}=1$

Otherwise, you can use l'Hospital's rule

**Hweengee** · October 12th 2008, 03:24 AM

Hi

I think I've gotten the 1st limit worked out, thanks to moo and the rest, but for the second I'm stuck at

$\lim_{x\to 0}~{\frac {\ln \left( 1+x+{x}^{2} \right) -x}{{x}^{2}}}$

Any suggestions on how to carry on? Or should I just use L'Hopital's rule?

**galactus** · October 16th 2008, 12:05 PM

You could try using the series expansion:

$ln(1+x+x^{2})=x+\frac{x^{2}}{2}-\frac{2x^{3}}{3}+\frac{x^{4}}{4}+\frac{x^{5}}{5}-\frac{x^{6}}{3}+..........$

Subtract off the x and we are left with:

$ln(1+x+x^{2})=\frac{x^{2}}{2}-\frac{2x^{3}}{3}+\frac{x^{4}}{4}+\frac{x^{5}}{5}-\frac{x^{6}}{3}+.........$

Now, divide by x^2 and get:

$\frac{1}{2}-\frac{2x}{3}+\frac{x^{2}}{4}+\frac{x^{3}}{5}-\frac{x^{4}}{3}+...........$

Now, take the limit as x-->0. See what it is?. Easy to see now.

**Hweengee** · October 17th 2008, 06:30 AM

yup. thanks for the help, much appreciated.

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