For 1, you cannot apply the rule because if you plug in negative infinity into x you would get 0
correct?
for 2, you plug in 0+ for x and you would get 1/infinity -0 and get 1/infinity
so it would be infinity? and thus you cannot apply the rule?
Use L'Hospitals Rule, if applicable to 1) lim x-->-infinity xe^x 2) lim x-->0+ 1/x -1/sqrtx?
2014-07-27 12:43 am
回答 (2)
2014-07-27 2:25 am
1) lim(x→ -∞) xe^x; this is of the form ∞ * 0
= lim(x→ -∞) x/e^(-x), now of the form ∞/∞
= lim(x→ -∞) 1/-e^(-x), by L'Hopital's Rule
= lim(x→ -∞) -e^x
= 0.
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2) lim(x→0+) (1/x - 1/√x); this is of the form ∞ - ∞
= lim(x→0+) (1/x - √x/x), common denominator
= lim(x→0+) (1 - √x)/x
= ∞, since (1 - √x)→1, but x→0 (from the positive side)
(L'Hopital's Rule is not applicable for this problem.)
I hope this helps!
= lim(x→ -∞) x/e^(-x), now of the form ∞/∞
= lim(x→ -∞) 1/-e^(-x), by L'Hopital's Rule
= lim(x→ -∞) -e^x
= 0.
----
2) lim(x→0+) (1/x - 1/√x); this is of the form ∞ - ∞
= lim(x→0+) (1/x - √x/x), common denominator
= lim(x→0+) (1 - √x)/x
= ∞, since (1 - √x)→1, but x→0 (from the positive side)
(L'Hopital's Rule is not applicable for this problem.)
I hope this helps!
2014-07-27 2:22 am
I did not do the problem, but try putting number 2 on a common denominator and then factoring out a x, afterwards input your limit and see if you get an indefinite form, if you do then you can apply l'hospitals rule.
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