# Math Help - [SOLVED] Exponential Limits

1. ## [SOLVED] Exponential Limits

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.

2. 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:

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

3. 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.

4. 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.

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

6. 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?

7. Originally Posted by RedBarchetta
How do you get from step 2 to step 3?
Do you know about asymptotic equivalences?

8. Originally Posted by Mathstud28
Do you know about asymptotic equivalences?
No.......? I'm halfway through Calculus II.

9. 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?

10. 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.

11. 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.

12. 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

13. 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?

14. 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.

15. yup. thanks for the help, much appreciated.