LIMIT question needs your attention

[khurram usman]

Limit x-->0- [[sqrt(88*x^2)] / x]
i think the answer of this question should be sqrt(88) but according to my instructor its -sqrt(88)
so please solve this one for me. only the left hand limit .

 

[LCKurtz]

Remember that sqrt(x²)= |x| = -x if x < 0.

 

[khurram usman]

Remember that sqrt(x²)= |x| = -x if x < 0.

ok...so the x in numerator will be positive and the one in the denominator negative.
so now the answer will be -sqrt(88)

 

[LCKurtz]

ok...so the x in numerator will be positive and the one in the denominator negative.
so now the answer will be -sqrt(88)

Not quite. The x in the numerator is equal to the x in the denominator; it's not that one of them is negative and the other positive. The point is that the fraction |x|/x = (-x)/x. Still, x itself is negative as x → 0^-.

 

[khurram usman]

i have one more question for you
cant we cut the two 'x' in the numerator and denominator without putting values?
if we do so then we won't get the - sign in the answer

 

[LCKurtz]

i have one more question for you
cant we cut the two 'x' in the numerator and denominator without putting values?
if we do so then we won't get the - sign in the answer

You can't cancel the x values in |x|/x without distinguishing the cases whether x > 0 or x < 0.

 

[khurram usman]

Not quite. The x in the numerator is equal to the x in the denominator; it's not that one of them is negative and the other positive. The point is that the fraction |x|/x = (-x)/x. Still, x itself is negative as x → 0^-.

sorry.i get your point now. but your last point has confused them in another problem:
lim x->5 [3x-15/(sqrt{x^2 + 25 - 10x}]
how will u solve this question for the left and right hand limits separately?

 

[LCKurtz]

sorry.i get your point now. but your last point has confused them in another problem:
lim x->5 [3x-15/(sqrt{x^2 + 25 - 10x}]
how will u solve this question for the left and right hand limits separately?

Given our previous example, what do you get for your denominator $\sqrt{(x-5)^2}$ ? It is the same idea.

 

[khurram usman]

Given our previous example, what do you get for your denominator $\sqrt{(x-5)^2}$ ? It is the same idea.

here is what i understand:
for right hand limit : 3(x-5)/|x-5|
and since both numerator and denominator are positive we get 3

for left handed : 3(x-5)/|x-5|
in this case the numerator becomes negative but the denominator stays positive . so we get -3
ryt?
and overall the limit of function doesn't exist at 3

 

[LCKurtz]

Yes that's right. Sorry for the delay getting back to you but the rest of my life intervened.

 

[khurram usman]

thanks for your help...you cleared my concept