use l'hospital's rule to find the limit:
1. lim (tan 2x)^x
x-> 0+
i got 1 using ln y, but im not sure if its right..
2. lim (1+ (a/x))^bx
x-> INFINITY
again i got 1...but am unsure
any help is appreciated...thanks!!
Hello,
I get 1 too
Substitute $\displaystyle t=\frac xa$2. lim (1+ (a/x))^bx
x-> INFINITY
Thus $\displaystyle bx=abt$ and $\displaystyle \frac ax=\frac 1t$
$\displaystyle \lim_{x \to \infty} \left(1+\frac ax\right)^{bx}=\lim_{t \to \infty} \left(1+\frac 1t\right)^{t \cdot (ab)}$
Remember the rule $\displaystyle (a^b)^c=(a^c)^b=a^{bc}$
Therefore :
$\displaystyle \lim_{t \to \infty} \left(1+\frac 1t\right)^{t \cdot (ab)}=\lim_{t \to +\infty} \left(\left(1+\frac 1t\right)^t\right)^{ab}=e^{ab} \quad (*)$
$\displaystyle (*) \lim_{t \to + \infty} \left(1+\frac 1t\right)^t=e$ (by taking the logarithm and applying l'Hospital's rule)
(a little substitution can be done if t tends to $\displaystyle -\infty$)
Otherwise, applying the logarithm :
$\displaystyle \ln(y)=\lim_{x \to \infty} bx \left(1+\frac ax\right)$
Then apply l'Hospital's rule...