# Limits

• Apr 11th 2013, 07:48 PM
Mr Rayon
Limits
hi guys,

Just started the topic on limits. I was just wondering whether somebody could help me go through the following questions.

Evaluate the following limits:

(a)$\displaystyle \lim_{x\rightarrow \infty } \frac{2x(x^2-1)}{x^2(5x-4)}$

(b)$\displaystyle \lim_{x\rightarrow \infty } \frac{x^3}{2x^2+7}$

(c)$\displaystyle \lim_{x\rightarrow -\infty } \frac{2x-1}{x^4+3x+1}$

Any help would be appreciated!
• Apr 11th 2013, 08:25 PM
ibdutt
Re: Limits
• Apr 11th 2013, 08:32 PM
Mr Rayon
Re: Limits
Quote:

Originally Posted by ibdutt

a) as the highest powers in both the numerator and denominator are the same when expanded, the limit is just the coefficients of the highest powers in the numerator and denominator.
So, 2/10 which simplifies to 1/5.

b) as the numerator's highest power is greater than the denominator's highest power - the limit does not exist.

c) as the numerator's power is lower than the denominator's power, the limit is equal to 0.

??

Is this correct?

• Apr 11th 2013, 09:21 PM
ibdutt
Re: Limits
You can verify by dividing the numerator and denominator by the highest power of the variable.
Q 1. it is x^3. On dividing the numerator and the denominator by x^3 we get [2 - 2/x^2 ] / [ 5 - 4/x] the limit of this as x approaches infinity is 2/5
Similarly you can find the limits of other questions.
• Apr 12th 2013, 08:36 AM
Petrus
Re: Limits
Quote:

Originally Posted by Mr Rayon
a) as the highest powers in both the numerator and denominator are the same when expanded, the limit is just the coefficients of the highest powers in the numerator and denominator.
So, 2/10 which simplifies to 1/5.

b) as the numerator's highest power is greater than the denominator's highest power - the limit does not exist.

c) as the numerator's power is lower than the denominator's power, the limit is equal to 0.

??

Is this correct?

Divide your b by $\displaystyle x^2$ in top and bottom