1 Attachment(s)
limits at negative infinity
Attachment 25349
when taking the limit as x goes to negative infinity, and we divided by negative sqrt(x^2) why does the term 5x not become -5?
the negative is associated only with the second term in the denominator but not the 5, why?
also why is the denominator not divided by a negative?
Re: limits at negative infinity
This is what happens:
}}{x \left(\frac{4}{x}+1\right)})
})
Suppose 
then 
thus
}{x\left (\frac{4}{x}+1\right)} = \lim_{x \to +\infty} \frac{5+\sqrt{1+\frac{2}{x}}}{\frac{4}{x}+1} = 6)
Suppose 
then 
Now (check it) the limit will be equal to 4
Re: limits at negative infinity
Perphas this also can help you: substituting for example 
in 
we get +\sqrt{(-3)^2+2(-3)}}{4+(-3)})
. Now, divide numerator by ^2}\;(=-3))
and denonimator by 
. What do you obtain? (substtitute now by 
)
