$${\frac{\left({\mathtt{3}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{4}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{3}}}{\mathtt{\,-\,}}{\mathtt{36}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}\right)}{\left({{\mathtt{x}}}^{{\mathtt{4}}}{\mathtt{\,-\,}}{\mathtt{25}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{144}}\right)}}$$

a.) Evaluate Lim x->infinity f(x) and lim x->negative infinity f(x) and then identify any horizontal asymptotes.

b.) Find the vertical asymptotes. For each vertical asymptote x = a, evaluate lim x->a- f(x) and lim x->a+ f(x).

I feel really stupid for not being able to solve this but I can't find b. someone help please?

Guest Feb 9, 2015

#1**+5 **

a). As x goes to +infinity, the first term in the numerator gets much bigger than the others, so onlyu 3x^{4} needs to be kept. The same reasoning applies to the denominator, where only x^{4} needs to be kept. So you are left with 3x^{4}/x^{4} or just 3. Similar reasoning applies as x goes to -infinity, and, since x^{4} is positive when x is negative, the limit is also 3 in this case.

b). The vertical asymptotes occur when the denominator is zero. Notice that the denominator can be factored as (x^{2} - 9)(x^{2} - 16) = (x + 3)(x - 3)(x + 4)(x - 4).

However, the numerator can also be factored as 3x^{2}(x+4)(x-3), so the whole expression becomes

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

This means there are vertical asymptotes at x = -3 and x = 4.

The graph below should help you determine the sign of the asymptotes as they are approached from each side.

.

Alan
Feb 9, 2015

#1**+5 **

Best Answer

a). As x goes to +infinity, the first term in the numerator gets much bigger than the others, so onlyu 3x^{4} needs to be kept. The same reasoning applies to the denominator, where only x^{4} needs to be kept. So you are left with 3x^{4}/x^{4} or just 3. Similar reasoning applies as x goes to -infinity, and, since x^{4} is positive when x is negative, the limit is also 3 in this case.

b). The vertical asymptotes occur when the denominator is zero. Notice that the denominator can be factored as (x^{2} - 9)(x^{2} - 16) = (x + 3)(x - 3)(x + 4)(x - 4).

However, the numerator can also be factored as 3x^{2}(x+4)(x-3), so the whole expression becomes

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

This means there are vertical asymptotes at x = -3 and x = 4.

The graph below should help you determine the sign of the asymptotes as they are approached from each side.

.

Alan
Feb 9, 2015