# Limits question

• Apr 21st 2008, 08:43 AM
akhayoon
Limits question
Limit question on a midterm that I have and still haven't the slightest idea on how to complete (Speechless)

Find all Values of a and b such that $\lim{x}\rightarrow{0}\frac{3sin(2x)+ax^{2}-bx}{x^2}=1$

all I could figure out was that if you use L'Hopitale's rule you would get

$\lim{x}\rightarrow{0}\frac{6cos(2x)+2ax+b}{2x}=1$

but I have no idea how to use this to solve for any variables

thank you
• Apr 21st 2008, 08:46 AM
Moo
Hello,

Why not using a second time l'Hôpital's rule ?
• Apr 21st 2008, 08:54 AM
akhayoon
because it would be (0/0) ?
• Apr 21st 2008, 08:57 AM
Moo
Hm ok, I just don't know correctly the conditions for applying this rule :D
• Apr 21st 2008, 09:00 AM
flyingsquirrel
Hi

Another approach : $\lim_{x\to 0} \frac{\sin(2x)-\sin(0)}{2x-0}=\cos 0 = 1$
Hence, $\lim_{x \to 0} \frac{3}{x}\frac{\sin(2x)}{x}+a-\frac{b}{x}=\lim_{x \to 0} \frac{6}{x} \frac{\sin(2x)}{2x}+a-\frac{b}{x}=\lim_{x\to 0}\frac{6-b}{x}+a=1$
When is this possible ?
• Apr 21st 2008, 11:58 AM
Mathstud28
Quote:

Originally Posted by akhayoon
Limit question on a midterm that I have and still haven't the slightest idea on how to complete (Speechless)

Find all Values of a and b such that $\lim{x}\rightarrow{0}\frac{3sin(2x)+ax^{2}-bx}{x^2}=1$

all I could figure out was that if you use L'Hopitale's rule you would get

$\lim{x}\rightarrow{0}\frac{6cos(2x)+2ax+b}{2x}=1$

but I have no idea how to use this to solve for any variables

thank you

L'hopital it again if you wish...it works...but look...you have two equations with two variables...solve for them using simulataneous equations
• Apr 21st 2008, 04:05 PM
mr fantastic
Quote:

Originally Posted by akhayoon
Limit question on a midterm that I have and still haven't the slightest idea on how to complete (Speechless)

Find all Values of a and b such that $\lim{x}\rightarrow{0}\frac{3sin(2x)+ax^{2}-bx}{x^2}=1$

all I could figure out was that if you use L'Hopitale's rule you would get

$\lim{x}\rightarrow{0}\frac{6cos(2x)+2ax+b}{2x}=1$

but I have no idea how to use this to solve for any variables

thank you

Obviously you require another indeterminant form. This will happen if b = -6. So then you have:

$\lim_{x \rightarrow 0} \frac{6 \cos(2x) + 2ax - 6}{2x}=1$.

Apply l'Hopital again:

$\lim_{x \rightarrow 0}\frac{-12 \sin (2x) + 2a}{2}=1$.

In other words, $\frac{2a}{2} = 1 \Rightarrow a = 1$.
• Apr 21st 2008, 04:15 PM
mr fantastic
Quote:

Originally Posted by akhayoon
Limit question on a midterm that I have and still haven't the slightest idea on how to complete (Speechless)

Find all Values of a and b such that $\lim{x}\rightarrow{0}\frac{3sin(2x)+ax^{2}-bx}{x^2}=1$

all I could figure out was that if you use L'Hopitale's rule you would get

$\lim{x}\rightarrow{0}\frac{6cos(2x)+2ax+b}{2x}=1$

but I have no idea how to use this to solve for any variables

thank you

All your troubles would have been avoided of course if you hadn't run to the l'Hospital with a sore thumb. Try substituting the series expansion around x = 0 for sin(2x):

$\Rightarrow \lim_{x \rightarrow 0} \frac{3 (2x - [2x]^3/6 + ...) + ax^{2} - bx}{x^2}=1$

$\Rightarrow \lim_{x \rightarrow 0} \frac{(6x - 4x^3/3 + ...) + ax^{2} - bx}{x^2}=1$

$\Rightarrow \lim_{x \rightarrow 0} \frac{(6 - b) x + ax^{2} - 4x^3/3 + ...}{x^2}=1$

$\Rightarrow \lim_{x \rightarrow 0} \frac{(6 - b) + ax - 4x^2/3 + ...}{x} = 1$

There's a problem unless 6 - b = 0 => b = 6. Then:

$\Rightarrow \lim_{x \rightarrow 0} \frac{ax - 4x^2/3 + ...}{x} = 1$

$\Rightarrow \lim_{x \rightarrow 0} \left(a - \frac{4x}{3} + ...\right) = 1$

Therefore a = 1.

(Cool)