Find the limit of the function algebraically. x→−3 (x^2-9/x^3+3)

Mathematics

1 answer:

Naya [18.7K]4 years ago

3 0

Answer:

d) 0

Step-by-step explanation:

The function is defined for x=-3, so it can simply be evaluated there.

$\displaystyle\lim_{x \to -3}\left(\frac{x^2-9}{x^3+3}\right)=\frac{(-3)^2-9}{(-3)^3+3}=\frac{9-9}{-27+3}=0$

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Step-by-step explanation:

The following laws of exponent are useful to this problem:

$\displaystyle \large{ {(mn)}^{b} = {m}^{b} {n}^{b} } \\ \displaystyle \large{ {n}^{ - b} = \frac{1}{ {n}^{b} } }$

We are given the expression:

$\displaystyle \large{( {m}^{8} {n}^{ - 4} )^{ \frac{1}{4} } }$

Use the first law of exponent above.

$\displaystyle \large{( {m}^{8} {n}^{ - 4} )^{ \frac{1}{4} } = {m}^{(8)( \frac{1}{4}) } {n}^{ ( - 4)( \frac{1}{4}) } } \\ \displaystyle \large{( {m}^{8} {n}^{ - 4} )^{ \frac{1}{4} } = {m}^{2 } {n}^{ - 1} }$

Make sure to recall the important necesscary fundamental math such as operation with negative numbers/integers, basic division, fraction, etc.

From the expression, apply the second law of exponent to n^-1.

$\displaystyle \large{( {m}^{8} {n}^{ - 4} )^{ \frac{1}{4} } = {m}^{2 } ( \frac{1}{ {n}^{1} } )} \\ \displaystyle \large{( {m}^{8} {n}^{ - 4} )^{ \frac{1}{4} } = {m}^{2 } ( \frac{1}{ {n} } )} \\$