Question

How do you find the limit of `(sqrt(x^2+1))/(3x-1)` as x approaches infinity?

Answer 1

The answer are:

for `xrarr+oo`, the limit is `1/3`,

For `xrarr-oo`, the limit is `-1/3`.

This is because:

`lim_(xrarr+-oo)sqrt(x^2+1)/(3x-1)=lim_(xrarr+-oo)sqrt(x^2)/(3x)=`

`=lim_(xrarr+-oo)|x|/(3x)=(1)`

For `xrarr+oo`, `|x|=x`,

for `xrarr-oo`, `|x|=-x`.

So:

`(1)=lim_(xrarr+oo)x/(3x)=1/3`,

`(1)=lim_(xrarr-oo)(-x)/(3x)=-1/3`.

Answer 2

We can easily find the limit by dividing every term by `x` as follows
Remember when we put x into the radical we must square it
I am assuming positive infinity. In other words x greater than or equal to zero.

`(sqrt{x^2/x^2+1/x^2})/((3x)/x-1/x)`

`(sqrt{1+1/x^2})/(3-1/x)`

Now taking the limit as x approaches infinity we have

`(sqrt{1+0})/(3-0)=sqrt{1}/3=1/3`