Question

Find a reasonable estimate of the limit

Answer 1

Answer:

The answer is (c) ⇒ the value is 6.6667

Step-by-step explanation:

∵ $\lim_{x\to \2} _2\frac{x^{5}-32}{x^{3}-8}$

∵ 32 = 2^5 , 8 = 2³

∴ $\lim_{x \to \2}_2 \frac{x^{5}-2^{5}}{x^{3}-2^{3} }$

* by using the rule:

$\lim_{x\to\a}_a \frac{x^{n}-a^{n}}{x^{m}-a^{m}}=\frac{n}{m}(a)^{n-m}$

∴ $\frac{5}{3}(2)^{5-3}=\frac{5}{3}(2)^{2}=\frac{20}{3}$

∴ 20/3 = 6.6667 ⇒ answer (c)

Answer 2

Answer:

The correct option is c.

Step-by-step explanation:

The given limit is

$lim_{x\rightarrow 2}\frac{x^5-32}{x^3-8}$

It is can be written as

$lim_{x\rightarrow 2}\frac{x^5-2^5}{x^3-2^3}$

According to the property of limits,

$lim_{x\rightarrow a}\frac{x^n-a^n}{x^m-a^m}=\frac{n}{m}(a)^{n-m}$

In the given limit, a=2, n=5 and m=3. Using the above property of limits we get

$lim_{x\rightarrow 2}\frac{x^5-2^5}{x^3-2^3}=\frac{5}{3}(2)^{5-3}$

$lim_{x\rightarrow 2}\frac{x^5-2^5}{x^3-2^3}=\frac{5}{3}(2)^{2}$

$lim_{x\rightarrow 2}\frac{x^5-2^5}{x^3-2^3}=\frac{5}{3}(4)$

$lim_{x\rightarrow 2}\frac{x^5-2^5}{x^3-2^3}=\frac{20}{3}$

$lim_{x\rightarrow 2}\frac{x^5-2^5}{x^3-2^3}=6.6667$

Therefore the correct option is c.