Question

\[\large\lim_{x\to0^+}\frac{\sqrt x\sin(3x)+x^2+\arctan(5x)}{\sin\Big(\arctan\big(\sin(x)\big)\Big)}\]

Answer 1

have you tried plugging in 0 yet?

Answer 2

That will not work actually ... Mira, don't u think it will lead to 0/0 form ?

Answer 3

\[\large\leadsto \tfrac00\]

Answer 4

yes it is 0/0
we plug in so that we know its of that form
and so we can use lhospitals rule because it is 0/0

Answer 5

taking derivatives might get messy

Answer 6

yes but thats the only way to proceed

Answer 7

plit it into 3 term atleast who knows some of them might end up going to inf if u are lucky and u can stop

Answer 8

we have to use lhospitals rule here
we just need to get the derivative of the top and the bottom
that's what lhospitals rule is about
we can use it for 0/0
or inf/inf
for those cases

Answer 9

What derivatives do You get ?

Answer 10

\(\lim_{x\rightarrow 0^+}\left(\cfrac{\cfrac{\sqrt{x} \sin (3x)}{\sqrt{1 - x \sin^2 (3x) }} + \cfrac{x^2}{\sqrt{1-x^4}} + \cfrac{\tan^{-1}(5x)}{\sqrt{1-\tan^{-1}(5x)^2}}}{\sin x}\right)\)

Answer 11

\[\leadsto\frac00\]

Answer 12

1 more lhopital gogo

Answer 13

Yeah.. its not complete yet!

Answer 14

ganeshie do u like my bebe

Answer 15

Dan, LH rule is not going to satisfy us. I will use something different here :-)

Answer 16

i was wondering why we don split that equation into 3 separate terms

Answer 17

and apply them separetly

Answer 18

http://www.wolframalpha.com/input/?i=%5Cfrac%7B%5Csqrt+x%5Csin%283x%29%2Bx%5E2%2B%5Carctan%285x%29%7D%7B%5Csin%28%5Carctan%28%5Csin%28x%29%7D
reverse engineer xD

Answer 19

do you want to see a simpler method?

Answer 20

yes!

Answer 21

No..

Answer 22

gimme hint first

Answer 23

I'm trying.. please..

Unkle, PM the hint to dan.

Answer 24

Don't post it here :P I have already wasted 3 pages... I want to do more work ...

Answer 25

/haha math

Answer 26

\[\large\lim_{x\to0^+}\frac{\sqrt x\sin(3x)+x^2+\arctan(5x)}{\sin\Big(\arctan\big(\sin(x)\big)\Big)}\]

I copied the question again so that I don't have to scroll again and again. :P

Answer 27

I really love that x^2 .. it looks the best! :P the others are nasty.

Answer 28

What if we expand each bit by Taylor series and cancel the high order terms to get 0+0+5=5

Answer 29

aw math xD

Answer 30

mathmale got it

Answer 31

That is what I tried... the other two are zero in my equation. The third expression should be 5.

Answer 32

the rest went to 0

Answer 33

@mathmate

\[\large\lim_{x\to0^+} \frac{\sqrt x\sin(3x)+x^2+\arctan(5x)}
{\sin\Big(\arctan\big(\sin(x)\big)\Big)}\]
\begin{align*}
&=\lim_{x\to0^+}\frac{\sqrt x\left[3x+O(x^3)\right]+x^2+\left[5x+O(x^3)\right]}
{\sin\Big(\arctan\big(\left[x+O(x^3)\right]\big)\Big)} \\
&=\lim_{x\to0^+}\frac{3x^{3/2}+O(x^{5/2})+x^2+5x+O(x^3)}
{\sin\Big(\left[x+O(x^3)\right]\Big)} \\
&=\lim_{x\to0^+}\frac{5x+3x^{3/2}+x^2+O(x^3)}
{\left[x+O(x^3)\right]} \\
&=\lim_{x\to0^+}\frac{x(5+3x^{1/2}+x+O(x^2))}
{x\big(1+O(x^2)\big)} \\
&=\lim_{x\to0^+}\frac{5+3x^{1/2}+x+O(x^2)}
{1+O(x^2)} \\
&=\frac51 \\
&=5
\end{align*}

Answer 34

\(\Large \sin (\arctan (\sin x)) \to \sin x \\ as \ x\to 0 \)

Answer 35

then just separate the denominator

Answer 36

and each ofthe 3 limits will be easy to solve

Answer 37

***separate the numerator

Answer 38

\(\Large \sin (\arctan (\sin x)) = \dfrac{\sin x}{\sqrt{1+\sin ^2 x}}\)

Answer 39

\(\sqrt{1+\sin^2 x} \to 1 ~ as x\to 0\)

Answer 40

sqrt x sin 3x /sin x -> 0
x^2/sin x -> 0
arctan 5x /sin x -> 5
as x-?0

Answer 41

\(\color{blue}{\text{Originally Posted by}}\) @mathslover
\(\lim_{x\rightarrow 0^+}\left(\cfrac{\cfrac{\sqrt{x} \sin (3x)}{\sqrt{1 - x \sin^2 (3x) }} + \cfrac{x^2}{\sqrt{1-x^4}} + \cfrac{\tan^{-1}(5x)}{\sqrt{1-\tan^{-1}(5x)^2}}}{\sin x}\right)\)
\(\color{blue}{\text{End of Quote}}\)

= 0 + 0 +5

Right!

Answer 42

@UnkleRhaukus I always read your problems with awe and had no clue whatsoever. I am so glad even to think about participating this time.

Answer 43

Taylor series is powerful,

Answer 44

I need to learn that *taylor series* .. it spoiled my work :/

Anyways, awesome question Unkle!!!

Answer 45

@mathslover, any function that is regular, that means that all its derivative exist and are continuous in an interval
can be expressed as a series of power