1. ## Limit, derivation, help please

Hi! I need some help in the following problems.

1. What is the limit if n tends to infinity:

[(n+2) choose 2]/(n choose 2)]^n

2. Derivation

y=(4^x-x^4)/log(7)x; that is the seven-based log of x

3. Derivation

y=ln[e^x+e^(-x)]

4. Limit if x tends to zero

1/[e^(2x)-1]-1/2x

Thanks a lot.

2. Originally Posted by csodacsiga
[(n+2) choose 2]/(n choose 2)]^n
Hint: $\displaystyle {a\choose 2} = \frac{a(a-1)}{2}$

2. Derivation

y=(4^x-x^4)/log(7)x; that is the seven-based log of x
This is the same as,
$\displaystyle y= \frac{e^{\ln 4 x} - x^4}{\left( \frac{\ln x}{\ln 7} \right)}$
Now apply the quotient rule.

y=ln[e^x+e^(-x)]
With chain rule we get,
$\displaystyle (e^x + e^{-x})' \cdot \frac{1}{e^x + e^{-x}} = \frac{e^x - e^{-x}}{e^x + e^{-x}}$

4. Limit if x tends to zero

1/[e^(2x)-1]-1/2x

Thanks a lot.
This one is not clear.

3. For 1., $\displaystyle {n+2\choose2}\bigg/{n\choose2} = \frac{\frac12(n+2)(n+1)}{\frac12n(n-1)} = 1+\frac{4n+2}{n(n-1)}$. So $\displaystyle \left({n+2\choose2}\bigg/{n\choose2}\right)^{\!\!n} = \left(1+\frac{4n+2}{n(n-1)}\right)^{\!\!n}$. This looks as

though it should be close to $\displaystyle \bigl(1+\tfrac4n\bigr)^n$, which has the limit $\displaystyle e^4$ as $\displaystyle n\to\infty$.

To justify that guess, you could take the logarithm of $\displaystyle \left(1+\frac{4n+2}{n(n-1)}\right)^{\!\!n}$ and show that it tends to 4 as $\displaystyle n\to\infty$.

4. Thank you both for the great helps.

1/[e^(2x)-1]-1/2x is

(one over (e to the power of two x) minus one) minus one over two x

5. Originally Posted by csodacsiga
Thank you both for the great helps.

1/[e^(2x)-1]-1/2x is

(one over (e to the power of two x) minus one) minus one over two x
$\displaystyle \frac1{e^{2x}-1} - \frac1{2x} = \frac{1+2x-e^{2x}}{2x(e^{2x}-1)}$. Now apply l'Hôpital two or three times.

6. I assume this is it:

$\displaystyle \lim_{x\to 0}\frac{1}{e^{2x}-1}-\frac{1}{2x}$?.

We can be lazy and use ol' L'Hopital.

Rewrite:

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

L'Hopital:

$\displaystyle \displaystyle{-\lim_{x\to 0}\frac{e^{2x}-1}{2xe^{2x}+e^{2x}-1}}$

L'Hopital again:

$\displaystyle \displaystyle{-\lim_{x\to 0}\frac{1}{2x+2}}$

Now, we can see it?.

Now, try it without L'Hopital. That is 'funner'.