Question

Limit Problem-to find the value of a and b.
Can anyone help me to do the problem?
http://i.imgur.com/rp4ifCl.png

Answer 1

It looks nasty. My suggestion would be to use l'Hospital's rule in a "reverse" manner.

See the way we usually use it is in a forward manner - we evaluate the limit of the numerator and the denominator, and then we apply l'Hospital's rule to find the limit (if it exists).

Here we know the limit exists, and we know that both the numerator and the denominator tend to 0. Suppose the numberator was f(x) and denominator was g(x) then:

lim of f(x)/g(x) as x->0 = lim of f'(x) / g'(x) as x->0 = 1

Answer 2

Please someone help.

Answer 3

I see no theoretical reason why it shouldn't work.

We press on then to find f'(x) and g'(x).

f'(x)=3*x^2
g'(x)=*oh gods*[ 1/( 2*sqrt(a+x) )] * [bx-sin(x)] + sqrt(a+x) *[b-cos(x)]

Answer 4

It's a pain in the retricemethod but once you get f'''(x) and g'''(x), the idea is that f'''(x)=1 and you will say that g'''(x) will also have to equal 1 for the limit itself to equal 1 and that should give you some information about a and b.

Again, it's a pelletty method but I see no way around this.

Answer 5

But it is becoming very lengthy

Answer 6

Do you know about power series?

Answer 7

I'll show you the method:

Answer 8

\[\Large\sf{\color{blueviolet}{\lim_{x\rightarrow0}\frac{x^3}{\sqrt{a+x}(bx-sinx)}\\=\lim_{x\rightarrow0}\frac{x^3}{\sqrt{a+x}\left(bx-\left(\frac{x}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}\ldots\right)\right)}}}\]

Answer 9

Now it's a matter of guessing what a and b should be for the limit to be 1.
I leave you to ponder over it....