Question 704028
That picture is somewhat difficult to read but you're interested in limit as x approaches positive and negative infinities for {{{f(x)=(2x-1)/(x^2+1)}}}.  Unless read incorrectly.


That function is defined for the entire real number line, so nomatter what x is picked, the function and limit are the same.  The denominator, x^2+1, is never zero.


The evaluation directly for limits at infinity may be difficult to see.  Try dividing numerator AND denominator by x^2:

{{{((2x-1)/x^2)/((x^2+1)/x^2)=((2/x)-1)/(1+1/x^2)}}}


Toward negative infinity, the numerator approaches 0-1 and denominator approaches 1+0.  The function then approaches -1/1, (negative 1).
... and you may then decide what happens as x approaches positive infinity?