Question

Use the limit theorem and the properties of limits to find the limit.

Answer 1

Answer:

Option C is right

Step-by-step explanation:

A function is given as

$h(x) = \frac{3x^2+5x+3}{8x^2-4x+5}$

Limit is to be found out for x tends to -infinity.

Since both are having even degrees when x tends to -infinity leading terms would turn positive.

We find that numerator and denominator has the same degree.

HEnce a horizontal asymptote exists

COefficients of leading terms are 3 and 8 respectively

Asymtote would be y =3/8

Alternate method:

When x tends to -infinity, 1/x tends to 0

$h(x) = \frac{3+\frac{5}{x}+\frac{3}{x^2}  }{\frac{8}{x^2} -\frac{4}{x}+5 }$[/tex]

by dividing both numerator and denominator by square of x.

Now take limit as 1/x tends to 0

we get

limit is y tends to 3/8

Hence horizontal asymptote is y =3/8

Answer 2

Answer:

c. 3/8

Step-by-step explanation:

Knowing that when x tends to -infinity, 1/x and 1/x^2 tends to 0, we have to divide both numerator and denominator by square of x ( because x^2, the major potency in the problem).

Now, we take limit and we have that:

lim x->inf $\frac{3   +5/x +3/x^2}{8 -4/x +5/x^2}$

we have that the terms where we have x and x^2 tends to zero, so the limit tends to 3/8.