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Calc Question

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

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
+26399
+5

a).  As x goes to +infinity, the first term in the numerator gets much bigger than the others, so onlyu 3x4 needs to be kept.  The same reasoning applies to the denominator, where only x4 needs to be kept.  So you are left with 3x4/x4 or just 3.   Similar reasoning applies as x goes to -infinity, and, since x4 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 (x2 - 9)(x2 - 16)  =  (x + 3)(x - 3)(x + 4)(x - 4).

However, the numerator can also be factored as 3x2(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
Sort:

#1
+26399
+5

a).  As x goes to +infinity, the first term in the numerator gets much bigger than the others, so onlyu 3x4 needs to be kept.  The same reasoning applies to the denominator, where only x4 needs to be kept.  So you are left with 3x4/x4 or just 3.   Similar reasoning applies as x goes to -infinity, and, since x4 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 (x2 - 9)(x2 - 16)  =  (x + 3)(x - 3)(x + 4)(x - 4).

However, the numerator can also be factored as 3x2(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
#2
0

Ah! I am so foolish, thank you so much!

Guest Feb 9, 2015

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