## kaiz122 3 years ago help me here. please

1. anonymous

$\lim_{x \rightarrow +\infty} (1+ \frac{k}{x})^{x}$

2. anonymous

any guesses?

3. anonymous

it's in indeterminate form $1^{\infty}$

4. anonymous

since on definition of $$e$$ is $$\lim_{x\to \infty}(1+\frac{1}{x})^x$$ it will be no surprise that $\lim_{x\to \infty}\left(1+\frac{k}{x}\right)^x=e^k$

5. anonymous

if that is an unacceptable answer, take the log, take the limit, which will be $$k$$ and then exponentiate to get $$e^k$$

6. anonymous

start with $x\ln(1+\frac{1}{x})$ which is now $$\infty\times 0$$ then the usual trick is to rewrite as $\frac{\ln(1+\frac{1}{x})}{\frac{1}{x}}$ so it is now $$\frac{0}{0}$$ and then use l'hopital

7. anonymous

let $y=(1+\frac{k}{x})^{x}$ $\ln{y} =x\ln(1+\frac{k}{x})$ then, $\lim_{x \rightarrow +\infty} x \ln (1+\frac{k}{x})$

8. anonymous

yup then go to $\frac{\ln(1+\frac{k}{x})}{\frac{1}{x}}$ and you should be good to go

9. anonymous

=0 *(+Infinity)

10. anonymous

right rewrite as i did above to get $$\frac{0}{0}$$

11. anonymous

we can rewrite it to this form $\lim_{x \rightarrow +\infty} \frac{\ln(1+ \frac{k}{x})}{\frac{1}{x}}$

12. anonymous

=$\frac{0}{0}$

13. anonymous

should we use the L' Hospital Rule now?

14. anonymous

yes

15. anonymous

what's the derivative of $\ln (1+\frac{k}{x})?$ is it $\frac{x^{2}}{(x+k)^{2}}?$

16. phi

no.

17. anonymous

ok so it's $\frac{-k}{x^{2}+kx}$

18. anonymous

u know 1^infinite will be =o

19. anonymous

i don't think it is.