Answer :
Let's solve each of the limits step by step.
1. Limit as [tex]\( x \)[/tex] approaches infinity for the first expression:
We have the expression:
[tex]\[
\lim_{x \to \infty} \frac{7x^9 - 4x^5 + 2x - 13}{3x^9 + x^8 - 5x^2 + 2x}
\][/tex]
- The highest power of [tex]\( x \)[/tex] in both the numerator and the denominator is [tex]\( x^9 \)[/tex].
- We divide every term in the numerator and the denominator by [tex]\( x^9 \)[/tex]:
[tex]\[
\lim_{x \to \infty} \frac{\frac{7x^9}{x^9} - \frac{4x^5}{x^9} + \frac{2x}{x^9} - \frac{13}{x^9}}{\frac{3x^9}{x^9} + \frac{x^8}{x^9} - \frac{5x^2}{x^9} + \frac{2x}{x^9}}
\][/tex]
- This simplifies to:
[tex]\[
\lim_{x \to \infty} \frac{7 - \frac{4}{x^4} + \frac{2}{x^8} - \frac{13}{x^9}}{3 + \frac{1}{x} - \frac{5}{x^7} + \frac{2}{x^8}}
\][/tex]
- As [tex]\( x \to \infty \)[/tex], terms involving [tex]\( \frac{1}{x^k} \)[/tex] where [tex]\( k > 0 \)[/tex] approach 0.
- Thus, the expression simplifies to:
[tex]\[
\frac{7}{3}
\][/tex]
So, the limit is [tex]\( \frac{7}{3} \)[/tex], which is approximately 2.3333.
2. Limit as [tex]\( x \)[/tex] approaches 4 for the second expression:
We have the expression:
[tex]\[
\lim_{x \to 4} \frac{x^4 - 256}{x^2 - 16}
\][/tex]
- Notice that [tex]\( x^4 - 256 \)[/tex] can be rewritten as a difference of squares: [tex]\((x^2 - 16)(x^2 + 16)\)[/tex].
- The denominator [tex]\( x^2 - 16 \)[/tex] can be factored as [tex]\((x - 4)(x + 4)\)[/tex].
- The original expression becomes:
[tex]\[
\lim_{x \to 4} \frac{(x^2 - 16)(x^2 + 16)}{(x - 4)(x + 4)}
\][/tex]
- Cancel the common factor [tex]\((x - 4)\)[/tex]:
[tex]\[
\lim_{x \to 4} \frac{x^2 + 16}{x + 4}
\][/tex]
- Now substitute [tex]\( x = 4 \)[/tex] into the simplified expression:
[tex]\[
\frac{4^2 + 16}{4 + 4} = \frac{16 + 16}{8} = \frac{32}{8} = 4
\][/tex]
So, the limit is 4.0.
In summary, the results are:
- The first limit is approximately 2.3333.
- The second limit is 4.0.
1. Limit as [tex]\( x \)[/tex] approaches infinity for the first expression:
We have the expression:
[tex]\[
\lim_{x \to \infty} \frac{7x^9 - 4x^5 + 2x - 13}{3x^9 + x^8 - 5x^2 + 2x}
\][/tex]
- The highest power of [tex]\( x \)[/tex] in both the numerator and the denominator is [tex]\( x^9 \)[/tex].
- We divide every term in the numerator and the denominator by [tex]\( x^9 \)[/tex]:
[tex]\[
\lim_{x \to \infty} \frac{\frac{7x^9}{x^9} - \frac{4x^5}{x^9} + \frac{2x}{x^9} - \frac{13}{x^9}}{\frac{3x^9}{x^9} + \frac{x^8}{x^9} - \frac{5x^2}{x^9} + \frac{2x}{x^9}}
\][/tex]
- This simplifies to:
[tex]\[
\lim_{x \to \infty} \frac{7 - \frac{4}{x^4} + \frac{2}{x^8} - \frac{13}{x^9}}{3 + \frac{1}{x} - \frac{5}{x^7} + \frac{2}{x^8}}
\][/tex]
- As [tex]\( x \to \infty \)[/tex], terms involving [tex]\( \frac{1}{x^k} \)[/tex] where [tex]\( k > 0 \)[/tex] approach 0.
- Thus, the expression simplifies to:
[tex]\[
\frac{7}{3}
\][/tex]
So, the limit is [tex]\( \frac{7}{3} \)[/tex], which is approximately 2.3333.
2. Limit as [tex]\( x \)[/tex] approaches 4 for the second expression:
We have the expression:
[tex]\[
\lim_{x \to 4} \frac{x^4 - 256}{x^2 - 16}
\][/tex]
- Notice that [tex]\( x^4 - 256 \)[/tex] can be rewritten as a difference of squares: [tex]\((x^2 - 16)(x^2 + 16)\)[/tex].
- The denominator [tex]\( x^2 - 16 \)[/tex] can be factored as [tex]\((x - 4)(x + 4)\)[/tex].
- The original expression becomes:
[tex]\[
\lim_{x \to 4} \frac{(x^2 - 16)(x^2 + 16)}{(x - 4)(x + 4)}
\][/tex]
- Cancel the common factor [tex]\((x - 4)\)[/tex]:
[tex]\[
\lim_{x \to 4} \frac{x^2 + 16}{x + 4}
\][/tex]
- Now substitute [tex]\( x = 4 \)[/tex] into the simplified expression:
[tex]\[
\frac{4^2 + 16}{4 + 4} = \frac{16 + 16}{8} = \frac{32}{8} = 4
\][/tex]
So, the limit is 4.0.
In summary, the results are:
- The first limit is approximately 2.3333.
- The second limit is 4.0.
