# Limits

• May 3rd 2011, 10:56 AM
prasum
Limits
lim x tending to infinity ((2-x)^40*(4+x)^5)/(2-x)^45
• May 3rd 2011, 11:06 AM
Plato
Quote:

Originally Posted by prasum
lim x tending to infinity ((2-x)^40*(4+x)^5)/(2-x)^45

HINT: $\frac{{\left( {2 - x} \right)^{40} \left( {4 + x} \right)^5 }}{{\left( {2 - x} \right)^{45} }} = \left( {\frac{{4 + x}}{{2 - x}}} \right)^5$
• May 3rd 2011, 11:09 AM
TheChaz
Quote:

Originally Posted by prasum
lim x tending to infinity ((2-x)^40*(4+x)^5)/(2-x)^45

Do you mean

$\lim_{x \to \infty} \frac{(2-x)^{40}(4+x)^5}{(2-x)^{45}}$

? There's a language for that.
• May 4th 2011, 05:36 PM
prasum
i am sorry in the numerator it is (2+x)^45
• May 4th 2011, 07:42 PM
HallsofIvy
Much the same. IF you were to multiply everything out, the numerator would be x^{45} (NOT -x^45) plus terms involving lower powers of x and the denominator would be -x^{45} plus terms involving lower powers of x.

General rule: If a rational function has polynomials of the same degree in numerator and denominator, the limit,as x goes infinity, is the ratio of those leading coefficients. If the numerator has lower degree than the denominator, the imit, as x goes to infinity, is 0. If the numerator has higher degree than the denominator, the limit, as x goes to infinity, does not exist.