Question

Another limit question.
Options :-
A) 9 p (ln 4)
B) 3 p (ln 4)^3
C) 12 p (ln 4)^3
D) 27 p (ln 4)^2

Answer 1

\[\huge \bf \lim_{x \rightarrow 0} \frac{(4^x-1)^3}{\sin(\frac{x}{p})\ln(1+\frac{x^2}{3})}\]

Answer 2

that one doesn't look easy to please
but when we plug in 0 we get 0/0
so my first thoughts are to do the l'hospital thing
and that will look ugly though

I will have to do a bit more thinking on this one

Answer 3

I would use Lhopitals

Answer 4

can you please show your next step using L HOPITAL's rule ?

Answer 5

look at options ! @freckles and @perl

Answer 6

$$
\LARGE \rm{
\lim_{x \rightarrow 0}~ \frac{(4^x-1)^3}{\sin(\frac{x}{p})\ln(1+\frac{x^2}{3})}
\\~\\\ \Rightarrow ^{Lhopitals}
\\
\lim_{x \rightarrow 0}~\frac{ 6(4^x-1)^24^x~ln(2)} { \frac{cos(\frac x p ) ln(1+\frac13 x^2)}{p}+\frac 23\frac{ sin(\frac xp )x}{(1+\frac 13 x^2)} }
}
$$

Answer 7

oh !! so lengthy !
don't think it will match the options !

Answer 8

i don't think this question will require L HOSTPITAL'S rule ? @perl and @freckles

Answer 9

any other simplest method ???

Answer 10

you can try to take the logarithm of the limit

Answer 11

suppose that limit is equal to L.
now find ln(L)

Answer 12

well you can cheat I think
I call this cheating anyways

set p=1
so your options are now
A) 9 (ln 4)
B) 3 (ln 4)^3
C) 12 (ln 4)^3
D) 27 (ln 4)^2

where your problem is now
\[\huge \bf \lim_{x \rightarrow 0} \frac{(4^x-1)^3}{\sin(x)\ln(1+\frac{x^2}{3})}\]
and plug in a number like .001
and see which of the choices are closer to that output


I don't like numerical methods for finding or approximating limits
but that is all I got right now

still trying to search for other options

Answer 13

you can also brute force the options.
plug in a value for p, and let x get very close to zero , say .0001.
which of the choices match

Answer 14

i don't know how to solve logarithm limit ?
I AM NEW LEARNER

Answer 15

well i thought wolfram would be able to handle it, it can't :(

Answer 16

@freckles you will surprise that in my grade maximum students will solve this question !
without putting p=1 and all that

Answer 17

it can handle it without the p

Answer 18

how ?

Answer 19

`i am in hurry` please type fast

Answer 20

I was talking about wolfram

Answer 21

oh !

Answer 22

i tried this , and it died
http://www.wolframalpha.com/input/?i=lim%28x-%3Einfinity%29+%284^x-1%29^3%2F%28+sin%28x%29*ln%281%2Bx^2%2F3%29%29

Answer 23

wait perl
it can do it with the p

Answer 24

then i have to consult this question by teacher or my friends !
and then i will definitely post the solution !

Answer 25

lol I thought it was goes to 0 not infinity

Answer 26

BBYYEE.,.. @freckles @perl
I HAVE TO GO NOW !!!

Answer 27

i get it is 8

Answer 28

with p = 1 , the limit is 8

Answer 29

@perl it x->0

Answer 30

http://www.wolframalpha.com/input/?i=lim+x-%3E0+%284%5Ex-1%29%5E3%2F%28sin%28x%2Fp%29*ln%281%2Bx%5E2%2F3%29%29

Answer 31

nevermind

Answer 32

3 ln (4)^3 is the answer

Answer 33

hey @perl did you get l'hospital to work?

Answer 34

like did you actually complete that way?

Answer 35

hey @perl you are correct

Answer 36

can you please explain ????

Answer 37

Applying Lhopitals once gave that more complicated expression.
I used the calculator ,
let .0001 -> A , store it

Answer 38

i put in x -> infinity , lol

Answer 39

in wolfram, that was my mistake

Answer 40

hmm, they say step by step solution, i wonder

Answer 41

opens wolfram mathematica 10

Answer 42

hey i think I have something interesting

Answer 43

\[\lim_{x \rightarrow 0}\frac{(\frac{4^x-1}{x})^3}{\frac{\sin(x)}{x} \cdot \frac{\ln(1+\frac{x^2}{3})}{x^2}}\]

Answer 44

I divided top and bottom by x^3

Answer 45

top goes to [ln(4)]^3

Answer 46

sin(x)/x->1

Answer 47

and that last thingy should go to (1/3)

Answer 48

\[\lim_{x \rightarrow 0}\frac{(\frac{4^x-1}{x})^3}{\frac{\sin(x)}{x} \cdot \frac{\ln(1+\frac{x^2}{3})}{x^2}}=\frac{(\ln(4))^3}{1 \cdot \frac{1}{3}}\]