Question

Help please..
when is indeterminate form used and where is it used ?

Answer 1

If you get \[\lim_{a \rightarrow b}\frac{ f(x) }{ g(x) } = \frac{ 0 }{ 0 }\]
or
\[\lim_{a \rightarrow b}\frac{ f(x) }{ g(x) } = \frac{ +/- \infty }{ +/-\infty }\]

Other cases are: (0)(+/- oo), \[1^\infty,0^0, \infty^0, \infty - \infty\]

Answer 2

ohh is that the same thing for l'hospital's rule ?

Answer 3

\[\lim_{x \rightarrow 0}\frac{ sinx }{ x } = \frac{ 0 }{ 0 }\]
Hence, we can use L'hopital rule:
\[\[\lim_{x \rightarrow 0}\frac{ sinx }{ x } = \lim_{x \rightarrow 0}\frac{ cosx }{ 1 }= \frac{ 1 }{ 1 }=1\]

Answer 4

ohh

Answer 5

\[\lim_{x \rightarrow \infty}=\frac{ e^x }{ x^2 }=\frac{ \infty }{ \infty } = \frac{ e^x }{ 2x }\]
Then apply L'hopital rule again, \[\lim_{x \rightarrow \infty}\frac{ e^x }{ 2x }=\frac{ e^x }{ 2 }=\infty \]

Answer 6

Sometimes, the L'hopital rule will be applied twice.

Answer 7

oh okay... can you help me with some problems

Answer 8

which ones?

Answer 9

one sec.

Answer 10

http://prnt.sc/awwj5d
i was confuse in todays lecture

Answer 11

part a?

Answer 12

yes a- c

Answer 13

what part are you confused on?

Answer 14

oh im confuse on everything.. hmm all i know is that i have to plugin the value of the limit to the equation

Answer 15

Since part(a) equation is a product, in order to use L'hopital rule -- the equation xlnx need to be rewritten as a quotient. L'hopital only work on quotients.

Answer 16

so you rewrite, xlnx as lnx / (1 / x)

Answer 17

so its like simplifying the equation ?

Answer 18

not, just rewriting the equation into a different form, just changing the structure of the equation.

Answer 19

\[f(x) = secxtanx\] for example, lets say I wanna use L'hopital -- but the equation isn't in the proper form to apply LH since it requires to be in fraction form.
So: \[f ( x ) = \frac{ tanx }{ cosx }\], now I can apply LH.

Answer 20

quetsion where did the tanx /cos x come from ?

Answer 21

\[secx = \frac{ 1 }{ cosx }\]
\[secxtanx = \frac{ 1 }{ cosx }* tanx = \frac{ tanx }{ cosx }\]

Answer 22

This is just an example of just rewriting the equation -- not related to your equations.

Answer 23

oh okay

Answer 24

then after applying LH:
\[\frac{ lnx }{ 1 / x } = \frac{ 1/x }{ -1/x^2 } = \lim_{x \rightarrow 0^+} - x\]

Answer 25

Looking at the graph of y = - x, as you approach from the right side of 0 -- the limit is zero.

Answer 26

Any questions on part(a)?

Answer 27

where did the -1/x^2 come from ?

Answer 28

the derivative of 1/x

Answer 29

quotient rule right ?

Answer 30

\[d/dx(\frac{ 1 }{ x }) = d/dx(x^{-1}) = (-1)x^{-2} = -\frac{ 1 }{ x^2 }\]

Answer 31

Power rule makes life easier sometimes

Answer 32

1/x is the same as x^-1

Answer 33

I'm just rewriting it, so I don't have to use the quotient rule. -- but you can use the quotient rule if you want. Either way works.

Answer 34

oh okayy

Answer 35

oh wait forgot ask something .. where did the 1/x come from for the denoniator ?

Answer 36

\[x \ln x = \frac{ lnx }{ \frac{ 1 }{ x } } <-> lnx * \frac{ x }{ 1 } = xlnx\]

Answer 37

oh so either one right ?

Answer 38

what do you mean either one?

Answer 39

Whatever the fraction rule is called. \[\frac{ a }{ \frac{ b }{ c } } = a * \frac{ c }{ b }\]

Answer 40

the one you wrote .. yeh

Answer 41

I am only changing x ln x into a fraction, just think of it as an equation changing forms.

Answer 42

oh okayy

Answer 43

but if its written this way .. does the 0 get pluggin right away ?