Question

Help me with L'hopital rule. lt x--->3
[(1/(x-3)) - (1/log(x-2))]

Answer 1

@mathmate plz help
@Psymon plz help its urgent

Answer 2

Well, the first step is to actually shift the function where it is in an undeterminant form, 0/0, infinity/infinity, etc.

Answer 3

its 0/0....What is the differentiation of log x?

Answer 4

Ah, it looks like it's just two separate terms, I apologize. Well, the derivative of log(base a)u is [1/[lna(u)] * du/dx

Answer 5

I suppose I should rewrite that better. [1/[(lna)*u]] * du/dx

Answer 6

To put it in 0/0 form, you can add the two fraction in the usual way.

Answer 7

\[\frac{1}{x-3} - \frac{1}{\ln (x-2)} = \frac{\ln(x-2)-(x-3)}{\ln(x-2) (x-3)}\]
now differentiate
\[\large \frac{\frac{1}{x-2} - 1}{\ln(x-2) + \frac{x-3}{x-2}}\]

Answer 8

I get
\( \frac{(log(x-2)-x+3)}{((x-3)log(x-2))} \)
Now you can apply l'Hôpital rule (twice) to get your answer

Answer 9

thanks a lot @mathmate @Psymon @dumbcow