Question

limits question...

Answer 1

\[\lim_{x \rightarrow 0}\frac{ \log(2+x) - \log(2-x) }{ x }\]

Answer 2

and there is one more ...
\[\lim_{x \rightarrow a}\frac{ asinx - xsina }{ ax^2 - xa^2 }\]

Answer 3

what methods are you spose to use to assess them?

Answer 4

Lhopital seems effective to this

Answer 5

i am not specifically aware of lhopital ?..but you can solve and may explain

Answer 6

take the ratio of the derivatives of the top and bottom

Answer 7

alright ..

Answer 8

\[\lim_{x \to 0}\frac{ \log(2+x) - \log(2-x) }{ x }=\lim_{x \to 0}\frac{ D_x[\log(2+x) - \log(2-x)] }{D_x[x] }\]
\[\lim_{x \to 0}\frac{ D_x[\log(2+x)] - D_x[\log(2-x)] }{D_x[x] }\]
\[\lim_{x \to 0}\frac{ \frac{1}{2+x} - \frac{1}{2-x} }{1}=0\]

Answer 9

Lhopital is useful for indeterminant stuff: 0/0 that is

Answer 10

in the x to a set up, a is held constant so treat it as a constant in the derivatives

Answer 11

well the answer key says the answer as 1

Answer 12

for the first one? if so, i may be recalling this a little off, or the key may be wrong :)

Answer 13

.... murmur, it is 1

Answer 14

doh! i derived the ln(2-x) as 1/(2-x) and dropped a negative :/

Answer 15

so yes, 1/2 + 1/2 = 1

Answer 16

\[\lim_{x \to 0}\frac{ \frac{1}{2+x} - \frac{-1}{2-x} }{1}=1\]
ahhh, thats better :)

Answer 17

yeah !!..true for the first...so buddy any idea for the second??

Answer 18

id do the same concept, but with proper derivatives

Answer 19

have you done derivatives yet?

Answer 20

well ..my tutor first didnt start with derivatives but i could solve the limit problems like by formulae ..sinx/x = 1 ...log(1 + x)/x = 1 ...and all that stuff!

Answer 21

you might want to split the fraction and factor the demoninator then

Answer 22

\[\lim_{x \rightarrow a}\frac{ asinx - xsina }{ ax^2 - xa^2 }\]
\[\lim_{x \rightarrow a}\frac{ asinx}{ ax^2 - xa^2 }+\frac{ xsina }{ ax^2 - xa^2 }\]
\[\lim_{x \rightarrow a}\frac{ sinx}{ x^2 - xa }+\frac{ sina }{ ax - a^2 }\]

Answer 23

yeah ..!..got it till here!

Answer 24

ln(x), ln(x+1), ln(x+2) are called horizontal shifts and have no effect on the limit.

Answer 25

can u plz solve it further too..

Answer 26

\[\lim_{x \rightarrow a}\frac{ sinx}{ x(x - a) }+\frac{ sina }{ a(x - a) }\]
i get a "dne" for this one

Answer 27

yeah i also did this

Answer 28

notice that the right side there simply cant be evaluated. its pretty much k/(x-a)

Answer 29

and the left part is 1/(x-a)

the limits simply fail to converge

Answer 30

wait the answer is given as (acosa - sina) / a^2

Answer 31

using Lhop
\[\lim_{x \rightarrow a}\frac{ asinx - xsina }{ ax^2 - xa^2 }\]
\[\lim_{x \rightarrow a}\frac{ acosx - sina }{ 2ax - a^2 }\]
\[\lim_{x \rightarrow a}\frac{ acos(a) - sina }{ 2a^2 - a^2 }\]
\[\frac{ acos(a) - sina }{ a^2}\]
maybe

Answer 32

oh, then it IS that :) Lhop is soo much simpler to figure out these things

Answer 33

yeah !!...i think dis method is nothing but diffrentiation ..right ??

Answer 34

yep

Answer 35

so long as your orginal limit is indeterminant; the Lhop is sooo useful

Answer 36

hmm..!..yeah !!..u r right ..just need to brush up that ...thank oyu very much to spend ur time for this ..:))

Answer 37

youre welcome :)

Answer 38

u indian?

Answer 39

no, floridian tho