# last question of the night..thank goodness

• Dec 17th 2009, 12:35 AM
twostep08
last question of the night..thank goodness
after hours of trying to make sense of this practice test, im at my last question...another doozy i have no idea on (you may have seen my plethora of other ones that i could not beat)

Use the
M - (delta)definition of limit at infinity to prove the following limits:
lim x>infinity... 3x^2/(5x^2 +1)=3/5
lim x> 1+...1/(x^2 -1)= infinity
• Dec 17th 2009, 12:45 AM
Drexel28
Quote:

Originally Posted by twostep08
after hours of trying to make sense of this practice test, im at my last question...another doozy i have no idea on (you may have seen my plethora of other ones that i could not beat)

Use the
M - (delta)definition of limit at infinity to prove the following limits:

lim x>infinity... 3x^2/(5x^2 +1)=3/5

lim x> 1+...1/(x^2 -1)= infinity

Do you know any theorems that you can use? Otherwise, note that $\displaystyle \left|\frac{3x^2}{5x^2+1}-\frac{3}{5}\right|=\left|\frac{3}{25x^2+5}\right|\ leqslant\frac{3}{x^2}$. So choosing $\displaystyle x$ such that $\displaystyle x>\frac{1}{\sqrt{\frac{3}{\varepsilon}}}$ ensures that $\displaystyle \left|\frac{3x^2}{5x^2+1}\right|\leqslant\frac{3}{ x^2}<\frac{3}{\sqrt{\frac{3}{\varepsilon}^2}}=\var epsilon$..................and done.