Calculus: Limit at Infinity?

x = ?x²

multiply through by 1/x †† -1/(?x²) on the bottom see the video in sources for the reason why††

(1/x)(-2x + 1) = -2 + (1/x)

-(1/?x²)?(x² + x) = -?(1 + (1/x))

plug in infinity

-2 + 0
---------
-?(1 + 0)


-2 / -1 = 2
 
1)note first that you have infinity/infinity undeterminate form
2) divide the top and the bottom by |x|=-x
that's a tricky problem because sqrt(x^2)=-x when x is negative, so this is why you divide by -x
 
lim (x -> -?) (-2x + 1)/ ?(x² + x)

since you have ?/? you can use L'hopitals

= lim (x -> -?) (-2)/ (x/?(x² + x))
= lim (x -> -?) -2?(x² + x) / x
= lim (x -> -?) -2?(x² + x) / -?x² (negative as we are considering negative x's)
= lim (x -> -?) 2?(x²/x² + x/x²)
= lim (x -> -?) 2?(1 + 1/x)
= 2
 
lim (-2x+1)/?(x^2 +x) is indeterminate (?/?).

So we use an algebraic manipulation to find the limit.

= lim [(-2x+1)/?(x^2 +x)] * [(1/?x^2) / (1/?x^2)]
= lim [(-2x+1) * (1/?x^2)] / [?(x^2 +x) * (1/?x^2)]
= lim [(-2x+1) * (1/|x|)] / [?(1 + 1/x)]

Since x approaches -?, |x| = -x.

= lim [(-2x+1) * (1/-x)] / [?(1 + 1/x)]
= lim [2-1/x] / [?(1 + 1/x)]
= 2 / ?1 = 2

The limit is 2.
 
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