Question

calculates the limit
\[\lim_{x \rightarrow 0 } \ln(x+1)/x\]

Answer 1

you know L'Hospitals rule?

Answer 2

on the other hand,,you know the shortcut formulla for 1^(infinity) types ?

Answer 3

yes i know l'hop rule
but how can i solve this limit

Answer 4

find the derivative of the numerator then find the derivative of the denominator

Answer 5

\[\lim_{x \rightarrow 0 } \frac{\ln(x+1)}{x}=\lim_{x \rightarrow 0 } \frac{\frac{d}{dx}[\ln(x+1)]}{\frac{d}{dx}[x]}\]

Answer 6

k

Answer 7

an alternative would have been
in cases like (1+f(x))^(g(x)) where f(x) tends to 0 and g(x) tends to infinity,
the ans is e^(f(x) * g(x) ) ..this thing can be derived easily..

here, you had log ( 1+x)^(1/x)
if you compare,its the same case..
so ans becomes
log(e^(x* 1/x) ) = 1