find the limit as x approaches infinity of (4x-7)/((5x^2)+3x+7)

Question

clara1223 · Answer

$$\lim_{x \rightarrow \infty}\frac{ 4x-7 }{ 5x ^{2}+3x+7 }$$

clara1223 · Answer

We haven't gone over limits as x approaches infinity in class before.

amistre64 · Answer

divide it all by the highest power of x to simplify.

anything left with an x in the denominator will zero out since 1/inf is a very very small amount of cake to get

clara1223 · Answer

so multiply the denominator by 1/x^2?

amistre64 · Answer

your basically searching for any horizontal asymptotes ... you covered them yet?

amistre64 · Answer

top and bottom yes

amistre64 · Answer

there are 'rules' which you can commit to memory if you have the gigabytes to play with :)

clara1223 · Answer

so I'm left with $$\frac{ \frac{ 4 }{ x }-\frac{ 7 }{ x ^{2} } }{ 5+\frac{ 3 }{ x }+\frac{ 7 }{ x ^{2} } }$$

amistre64 · Answer

now everything with an x under it, goes to something very very small,  they approach zero.  what are we left with?

clara1223 · Answer

1/5?

amistre64 · Answer

powers of x that is ... 

4/x - 7/x^2  is not 1

clara1223 · Answer

-1?

amistre64 · Answer

0-0 = ??

clara1223 · Answer

oh so the answer is 0?

amistre64 · Answer

yep

clara1223 · Answer

because 0/5=0

amistre64 · Answer

one rule, if the bottom is a higher degree than the top, as x approaches infinity, the limit is zero

amistre64 · Answer

if they are of the same degree, they limit to their leading coeffs .. its just simpler to divide off by the highest power of x and assess