Limit of Function Homework: Solving Challenging Examples | Let Our Experts Help

[WhatIsTheLim]

Homework Statement

hello, I have a little problem with some limit of the functions, could you help me? please

Homework Equations

\lim_{x \to 0} \quad \displaystyle\frac{\sqrt[3]{1+\mbox{arctg} 3x}-\sqrt[3]{1-\arcsin 3x}}{\sqrt{1-\arcsin 2x}-\sqrt{1+\mbox{arctg} 2x}}<br />

The Attempt at a Solution

it's one of the hardest example from my book, i don't even know how to start it.
i will be very grateful for help

and.. i don't know why my tex doesn't work. do you have some idea?

 

[Mark44]

Homework Statement

hello, I have a little problem with some limit of the functions, could you help me? please

Homework Equations

\lim_{x \to 0} \quad \displaystyle\frac{\sqrt[3]{1+\mbox{arctg} 3x}-\sqrt[3]{1-\arcsin 3x}}{\sqrt{1-\arcsin 2x}-\sqrt{1+\mbox{arctg} 2x}}

The Attempt at a Solution

it's one of the hardest example from my book, i don't even know how to start it.
i will be very grateful for help

and.. i don't know why my tex doesn't work. do you have some idea?
please correct me,

You were missing the [ tex] and [ /tex] tags (without the leading spaces inside the brackets).

 

[Mark44]

Try multiplying by 1 in the form of (1 + arctan(3x))^(2/3) + (1 + arctan(3x))^(1/3)(1 - arcsin(3x))^(1/3) + (1 - arcsin(3x))^(2/3) over itself.

The underlying idea is that (a - b)(a^2 + ab + b^2) = a^3 - b^3.

 

[WhatIsTheLim]

Try multiplying by 1 in the form of (1 + arctan(3x))^(2/3) + (1 + arctan(3x))^(1/3)(1 - arcsin(3x))^(1/3) + (1 - arcsin(3x))^(2/3) over itself.

The underlying idea is that (a - b)(a^2 + ab + b^2) = a^3 - b^3.

i did just like you said and i got " (0*2)/[0(1+0+0)] :( that means that this is equal 0? right?

i typed this: ((1+(arc tg3x))^1/3 -(1+(arc sin3x))^1/3)/((1-(arc sin2x))^1/2-(1+(arc tg2x))^1/2) into wolfram alpha and there is too 0, but even that i don't think that my calculation is good in mathematician meaning.

what can i do else, what i do wrong?
any idea?

 

[Mark44]

[0/0] is an indeterminate form, which means that an expression that has this form can have any limiting value. Give me a while to take a closer look at this.

 

[WhatIsTheLim]

\frac{\sqrt[3]{1+\arctan 3x}-\sqrt[3]{1-\arcsin 3x}}{\sqrt{1-\arcsin 2x}-\sqrt{1+\arctan 2x}}=\\=\frac{\arctan 3x+\arcsin 3x}{-\arcsin 2x-\arctan 2x} \cdot \frac{\sqrt{1-\arcsin 2x}+\sqrt{1+\arctan 2x}}{\sqrt[3]{(1+\arctan 3x)^2}+\sqrt[3]{(1+\arctan 3x)(1-\arcsin 3x)}+\sqrt[3]{(1-\arcsin 3x)^2}}

so we got -3/2 *2/3 and that equals -1? is that correct? i have to be sure

 

[Bohrok]

-1 is the correct limit.

 

[WhatIsTheLim]

yeap, i know that this is the correct limit, but is this a correct calculation :-> -3/2 *2/3->-1

 

[Mark44]

-3/2 * 2/3 = -1, yes. Is that what you're asking?

 

[Mark44]

Looks good.

 

[WhatIsTheLim]

-3/2 * 2/3 = -1, yes. Is that what you're asking?

by -3/2 i was meaning the first statement of the above equation... :]

but.. i was always thinking that arcus sin3x is going to 0, lim x->0 arcus tg is going to 0 too?
why here we doesn't the same : arcussin3x->0, arcustg3x->0 -->> 0/0 in first statement.

 

[Mark44]

Because 0/0 is not a number. It is one of several indeterminate forms, which means that a limit having this form can come out to be anything. For example, as x --> 0, x/(2x) --> 1/2, x^2/x --> infinity, and x/x^3 --> 0. All three of these limits are of the form [0/0] and their limiting values are different.