## kaiz122 Group Title help me here. please one year ago one year ago

1. kaiz122

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

2. satellite73

any guesses?

3. kaiz122

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

4. satellite73

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. satellite73

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

6. satellite73

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. kaiz122

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. satellite73

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

9. kaiz122

=0 *(+Infinity)

10. satellite73

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

11. kaiz122

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

12. kaiz122

=$\frac{0}{0}$

13. kaiz122

should we use the L' Hospital Rule now?

14. satellite73

yes

15. kaiz122

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

16. phi

no.

17. kaiz122

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

18. sunnymony

u know 1^infinite will be =o

19. kaiz122

i don't think it is.