Question

limit question:
how do i evaluate lim x-> 0+ [x/(ln(x))] if the numerator -> 0 while the denominator -> -inf

Answer 1

\[\large \lim_{x \rightarrow 0+} \frac{x}{\ln(x)}\]

Answer 2

ans is 0

Answer 3

i know what the answer is I want to know why it is zero

Answer 4

@estudier

Answer 5

diffenrtiate num and den....and apply the limit

Answer 6

or expand ln x

Answer 7

and since x is very small higher powers of x can be neglected

Answer 8

you cant differentiate the top and bottom and then apply limit (L'Hospital) as its not in 0/0 or infty/infty form.....can u?

Answer 9

take x common from both den and num

Answer 10

wait.....

Answer 11

write it ln x as e^ln(lnx).
We get x/e^ln(lnx). Denominator comes something less than 1 and numerator comes zero.
I hope this is correct.
Waisted 20-30 mins on it

Answer 12

This might have come too late, but anyway...
Just recall that
\[\lim_{x \rightarrow a}f(x)g(x)=\left( \lim_{x \rightarrow a}f(x) \right)\left( \lim_{x \rightarrow a}g(x) \right)\]
If both limits exist.
It turns out, here, in your case:
\[f(x) = x\]\[g(x)=\frac{1}{\ln x}\]That's a good place to start :)

Answer 13

You can use L'hopitals rule on this one. ie f'(x)/g'(x). The derivative of x is 1 and the derivative of ln(x) is 1/x. You can always use L'hopitals rule but you can use power series to evaluate and also use them to get a fundamental understanding of the limits of many transcendental. Unfortunately most calculus courses do not teach power series until well after limits so the understanding must be based on memorization unfortunately.

You can also use a graphing calculator in many instances to understand these functions.

Answer 14

oops You can always use should be..................You can't always use

Answer 15

lol.....I forgot what terenzreignz quoted...very true