Trigonometry limit

[dome123]

I have a problem solving this limit since the use of L'Hospitals rule is not allowed in this case:

I have tried multiplying with ( sqrt(1+xsinx) + cosx ) / ( sqrt(1+xsinx) +cosx ) but I can't get to a solution.

 

[Deleted member 4993]

I have a problem solving this limit since the use of L'Hospitals rule is not allowed in this case:

I have tried multiplying with ( sqrt(1+xsinx) + cosx ) / ( sqrt(1+xsinx) +cosx ) but I can't get to a solution.

What did you get after the multiplication.

Do you notice that you can factor out sin²(x) from the denominator?

 

[dome123]

Yes, the last step I am on is written this way:

lim sinx(sqrt(1(xsinx) +cosx) / (sinx +x )

 

[MarkFL]

Let's look at the result on the expression of what you tried:

[MATH]\frac{\sin^2(x)(\sqrt{1+x\sin(x)}+\cos(x))}{1+x\sin(x)-\cos^2(x)}[/MATH]
Or:

[MATH]\frac{\sin^2(x)(\sqrt{1+x\sin(x)}+\cos(x))}{1-\cos^2(x)+x\sin(x)}[/MATH]
Apply a Pythagorean identity:

[MATH]\frac{\sin^2(x)(\sqrt{1+x\sin(x)}+\cos(x))}{\sin^2(x)+x\sin(x)}[/MATH]
Divide numerator and denominator by \(\sin^2(x)\):

[MATH]\frac{\sqrt{1+x\sin(x)}+\cos(x)}{1+\dfrac{x}{\sin(x)}}[/MATH]
Can you proceed?

 

[Deleted member 4993]

Yes, the last step I am on is written this way:

lim sinx(sqrt(1(xsinx) +cosx) / (sinx +x )

lim sinx(sqrt(1+xsinx) +cosx) / (sinx +x )

= lim (sqrt(1+xsinx) +cosx) / (x/sinx +1 ) ............................ contunue........

 

[dome123]

Let's look at the result on the expression of what you tried:

[MATH]\frac{\sin^2(x)(\sqrt{1+x\sin(x)}+\cos(x))}{1+x\sin(x)-\cos^2(x)}[/MATH]
Or:

[MATH]\frac{\sin^2(x)(\sqrt{1+x\sin(x)}+\cos(x))}{1-\cos^2(x)+x\sin(x)}[/MATH]
Apply a Pythagorean identity:

[MATH]\frac{\sin^2(x)(\sqrt{1+x\sin(x)}+\cos(x))}{\sin^2(x)+x\sin(x)}[/MATH]
Divide numerator and denominator by \(\sin^2(x)\):

[MATH]\frac{\sqrt{1+x\sin(x)}+\cos(x)}{1+\dfrac{x}{\sin(x)}}[/MATH]
Can you proceed?

I am stuck on that step exactly.

 

[Deleted member 4993]

I am stuck on that step exactly.

What is:

\(\displaystyle Limit_{x\to 0}\left[\frac{x}{sin(x)}\right] = ?\)

 

[dome123]

I know that lim sinx/x = 1 when x goes zero. Is it the same when they swith places?

 

[Deleted member 4993]

I know that lim sinx/x = 1 when x goes zero. Is it the same when they swith places?

Yes - because [1/1] = 1

 

[MarkFL]

I know that lim sinx/x = 1 when x goes zero. Is it the same when they swith places?

Is it true that:

[MATH]\lim_{x\to c}\frac{1}{f(x)}=\frac{1}{\lim\limits_{x\to c}f(x)}[/MATH] ?

 

[dome123]

Is it true that:

[MATH]\lim_{x\to c}\frac{1}{f(x)}=\frac{1}{\lim\limits_{x\to c}f(x)}[/MATH] ?

Yes.

 

[dome123]

Thanks for the help guys.

 

[Steven G]

The lim as x->0 (sin(x)/x) = 1 SINCE sin(x) and x are very close to the same value when x is near 0. Therefore in the case of lim as x->0 (x/sin(x)) this limit will also be 1 as the numerator and denominator are approximately the same. Is that clear?

Suppose lim as x->x₀(f(x)/g(x)) = 2, than what would lim as x->x₀(g(x)/f(x)) equal? Can you explain why?

 

[Steven G]

Yes - because [1/1] = 1

You are learning. I think you are finally ready to tackle 2/2.