1. ## limit 2

http://mustafayagci.com.tr/PDF/MY_M3.pd

Page 156 question 30.
Find a.b

I tried to get x^2 out of the parantheses and tried to use L'hospital rule.

2. ## Re: limit 2

Originally Posted by kastamonu
http://mustafayagci.com.tr/PDF/MY_M3.pdf

Page 156 question 30.
Find a.b

I tried to get x^2 out of the parantheses and tried to use L'hospital rule.

3. ## Re: limit 2

\begin{align*}\displaystyle \lim_{x \to \infty} \left( \dfrac{x^2}{x^2+1} \dfrac{\dfrac{1}{x^2}}{\dfrac{1}{x^2}} + (a-3)x+b-1\right) & =5 \\ \left(\displaystyle \lim_{x \to \infty} \dfrac{1}{1+\cancel{\dfrac{1}{x^2}}^0} + \lim_{x\to \infty}(a-3)x\right) +b-1 & = 5 \\ 1+\lim_{x\to\infty}(a-3)x + b - 1 & = 5 \end{align*}

4. ## Re: limit 2

This is where I left. If I take x out of parantheses I can't find a. I have only one choice. I will take a=3. And b=5.

5. ## Re: limit 2

Originally Posted by kastamonu
This is where I left. If I take x out of parantheses I can't find a. I have only one choice. I will take a=3. And b=5.
That is correct: $3\cdot 5=15.\;\;\; \bf{A}$

6. ## Re: limit 2

$\displaystyle \lim_{x\to \infty}px$ for p anything other than 0 is "infinity" (or "does not exist"). In order that this have any finite limit, a- 3 must be 0 so a must be 3. In that case, the limit is 1- b- 1= 5 so that b= 5.

Many Thanks.