What did you get after the multiplication.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.
Yes, the last step I am on is written this way:
lim sinx(sqrt(1(xsinx) +cosx) / (sinx +x )
I am stuck on that step exactly.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?
What is:I am stuck on that step exactly.
Yes - because [1/1] = 1I know that lim sinx/x = 1 when x goes zero. Is it the same when they swith places?
I know that lim sinx/x = 1 when x goes zero. Is it the same when they swith places?
Yes.Is it true that:
[MATH]\lim_{x\to c}\frac{1}{f(x)}=\frac{1}{\lim\limits_{x\to c}f(x)}[/MATH] ?
You are learning. I think you are finally ready to tackle 2/2.Yes - because [1/1] = 1