Let f(x)=(x^2)/1-x. Determine (limx->1+) xf(x)

Question

how do I solve this? 

Answer 1

lim_x→1[1 + xf(x)] = lim_x→1[1 +x³/(1-x)]

 

                         = lim_x→1[(x³-x+1)/(1-x)]

 

As x→1⁺, the numerator approaches 1 and the denominator approaches 0 from the left, so

 

           lim_x→1⁺ [(x³-x+1)/(1-x)] = -∞

 

As x→1^-, the numerator approaches 1 and the denominator approaches 0 from the right, so

 

           lim_x→1^- [(x³-x+1)/(1-x)] = ∞

 

Therefore, lim_x→1(1+xf(x)) does not exist.  The graph of y = 1+xf(x) has a vertical asymptote at x = 1.  

Answer 2

lim       1 + x[x² / (1 - x)]

x-->1

 

 

lim      1 + (x³ / (1 - x))

x-->1

 

 

lim        [(1 - x) + x³] / (1 - x)

x-->1

 

 

lim       (x³ - x + 1) / (1 - x)

x-->1

 

 

If we plug in x=1, the limit becomes undefined.  So we need to multiply the limit by (1 + x) / (1 + x).

 

lim      [(x³ - x + 1)(1 + x)] / [(1 - x)(1 + x)]

x-->1

 

 

lim       [(x³ - x + 1)(1 + x)] / (1 + x²)

x-->1

 

 

Now we can plug in x=1.

 

lim       2 / 2

x-->1

 

lim         1

x-->1

 

 

As x approaches 1, the limit is 1.