How can I separate tan x and sin x in a limit problem?

[harman the destroyer]

Homework Statement

there is one question
limx--->0tanx-sinx/x3

i actually tried to seprate tanx and sinx amd then i multiplied and divided by tan2x and sin2x so that i can make tan3x/x3and sin3x/x3 to be 1 and in the end sin2x canceled and i got the answer as -1 which is wrong

what errror have i done here

i hope this is in homework section

Homework Equations

The Attempt at a Solution

 

[Eclair_de_XII]

Yes, you have the right idea in trying to separate $\lim_{x \rightarrow 0} \frac{tan(x)-sin(x)}{x^3}$ into two terms. But why don't you try to use L'Hopital's rule consecutively for each term?

I think it would work better than multiplying and dividing the whole equation by $tan(2x)$ and the like.

 

[haruspex]

tried to seprate tanx and sinx

Not a good idea. $\frac{\tan(x)}{x^3}$ and $\frac{\sin(x)}{x^3}$ will each tend to infinity and you will end up with ∞-∞.

 

[haruspex]

The simplest way is to use the power series expansions of sin and cos, but I don't know whether you are permitted that.

 

[Delta2]

Use 3 times the Le Hopital Rule in the original fraction. The denominator will become 6, and I believe you can handle what's in the numerator.

EDIT: Using only 2 times the rule might be enough as well.

EDIT2: If you are not allowed to use Le Hop rule, then by replacing tanx=sinx/cosx and doing some algebra you can endup with 2sinx-sin(2x) in the numerator (and 2x^3cosx in the denominator).

Wolfram says that
$2sinx-sin(2x)=8sin^3(\frac{x}{2})cos(\frac{x}{2}) (1)$

and I guess after that it can go easy since you ll have essentially $(sin(\frac{x}{2})/\frac{x}{2})^3$ limit but right now I can't see the trigonometry required to establish (1).

 

[harman the destroyer]

Yes, you have the right idea in trying to separate $\lim_{x \rightarrow 0} \frac{tan(x)-sin(x)}{x^3}$ into two terms. But why don't you try to use L'Hopital's rule consecutively for each term?

I think it would work better than multiplying and dividing the whole equation by $tan(2x)$ and the like.

Actually in my school exams we are not allowed to use l'hospital rule we can use it only in competitve exams

Not a good idea. $\frac{\tan(x)}{x^3}$ and $\frac{\sin(x)}{x^3}$ will each tend to infinity and you will end up with ∞-∞.

.
Let me try to make u understand

I am saying that i separated tanx term and sine x term

Now i multiplied and divided tan term with tan^2x and sin with si^2x

Then i combine tan^3x and limit comes out to be 1 and i am left with tan^2x in denominator and similar i have done with sine then i used 1-sin^2x =cos^2x property and got -1
Do it step by step as i have done
U will understand what i want to convey

The simplest way is to use the power series expansions of sin and cos, but I don't know whether you are permitted that.

No

 

[Delta2]

You are doing a subtle mistake. You do not have the right to recombine the separate limits (1/tan^2x) and (-1/sin^2x)

The summation rule for limits
$\lim f(x)+g(x)=\lim f(x)+\lim g(x)$
holds only if the separate limits exist and at least one of them is a real number or both are $+\infty$ or both are $-\infty$. Here one goes to $+\infty$ and the other to $-\infty$ so the summation rule can not apply.

 

[ehild]

@Delta²
factor out sin(x) in the numerator and bring the terms to common denominator: $\frac {\sin(x)} {\cos(x)} \frac{1-cos(x)}{x^3}$.
Use the half-angle formula to replace 1-cos(x) by 2 sin²(x/2). Now you can group the factors in form of sin(x)/x and
sin (x/2)/(x/2).

 

[harman the destroyer]

o

You are doing a subtle mistake. You do not have the right to recombine the separate limits (1/tan^2x) and (-1/sin^2x)

The limit rule for summation
$\lim f(x)+g(x)=\lim f(x)+\lim g(x)$
holds only if the separate limits exist and converge to real numbers or both to $+\infty$ or both to $-\infty$. Here one goes to $+\infty$ and the other to $-\infty$ so the summation rule can not apply.

h thanks i was unaware of this fact thanku so much

 

[Mark44]

then i multiplied and divided by tan2x and sin2x so that i can make tan3x/x3and sin3x/x3

The above is unreadable without context. tan2x and tan3x could be confused with $\tan(2x)$ and $\tan(3x)$. Since you apparently mean $\tan^2(x)$ and $\tan^3(x)$, at least write these as tan^2(x) and tan^3(x). Be sure to use parentheses. Also, use ^ to denote an exponent, as x^3 rather than x3.

 

[harman the destroyer]

i did but i don't know why it is not showing i will take care of this

 

[Ray Vickson]

Homework Statement

there is one question
limx--->0tanx-sinx/x3

i actually tried to seprate tanx and sinx amd then i multiplied and divided by tan2x and sin2x so that i can make tan3x/x3and sin3x/x3 to be 1 and in the end sin2x canceled and i got the answer as -1 which is wrong

what errror have i done here

i hope this is in homework section

Homework Equations

The Attempt at a Solution

You are writing
$$\lim_{x \to 0} \tan x - 3 \frac{\sin x}{x} $$
when we read your problem statement using standard parsing rules for mathematical expressions. Is that really what you mean?

You need parentheses, because "a-b/c" reads as $a - \frac{b}{c}$; if you want $\frac{a-b}{c}$ you need brackets: "(a-b)/c". Also, you need to distinguish between x3 (that is, $3x$) and x^3 (that is, $x^3$). More to the point, "a/x3" means $\frac{a}{x} 3 = 3 \frac{a}{x}$.

 

[Mark44]

The simplest way is to use the power series expansions of sin and cos

That's definitely not the simplest way, and probably isn't appropriate for someone just starting to learn about limits.

Use 3 times the Le Hopital Rule in the original fraction

As already noted, the OP has been told that L'Hopital is not allowed.

As an alternative to the advice given by @ehild in post #8, a different identity can be used.
$\frac{\tan(x) - \sin(x)}{x^3} = \frac{\frac{\sin(x)}{\cos(x)} - \sin(x)}{x^3} = \frac{\sin(x)(1 - \cos(x))}{x^3\cos(x)}$
Simply multiply num. and denom. by $1 + \cos(x)$. Another step gets you the product of two fractions whose limit can be taken without much effort.

 

[ehild]

As an alternative to the advice given by @ehild in post #8, a different identity can be used.
$\frac{\tan(x) - \sin(x)}{x^3} = \frac{\frac{\sin(x)}{\cos(x)} - \sin(x)}{x^3} = \frac{\sin(x)(1 - \cos(x))}{x^3\cos(x)}$
Simply multiply num. and denom. by $1 + \cos(x)$. Another step gets you the product of two fractions whose limit can be taken without much effort.

It is really the simplest way !