Question

lim. x^5-32÷x^3-8
x-2​

Answer 1

EXPLANATION.

$\sf \implies \lim_{x \to 2} \dfrac{x^{5} - 32}{x^{3} - 8}$

As we know that,

Put the value of x = 2 in equation and check their indeterminant form.

$\sf \implies \lim_{x \to 2} \dfrac{(2)^{5} - 32}{(2)^{3} - 8}$

$\sf \implies \lim_{x \to 2} \dfrac{32 - 32}{8 - 8}$

$\sf \implies \lim_{x \to 2}\dfrac{0}{0}$

As we can see that,

It is a form of 0/0 indeterminant form.

We can simply factorizes the equation, we get.

Formula of :

(x⁵ - y⁵) = (x - y)(x⁴ + a³b + a²b² + ab³ + b⁴).

(x³ - y³) = (x - y)(x² + xy + y²).

We can write as,

(x⁵ - 2⁵) = (x - 2)(x⁴ + x³(2) + x²(2)² + x(2)³ + (2)⁴).

(x⁵ - 2⁵) = (x - 2)(x⁴ + 2x³ + 4x² + 8x + 16).

(x³ - 2³) = (x - 2)(x² + 2x + 4).

Using the formula in equation, we get.

$\sf \implies \lim_{x \to 2} \dfrac{x^{5} - 2^{5} }{x^{3} - 2^{3} }$

$\sf \implies \lim_{x \to 2} \dfrac{(x - 2)(x^{4} + 2x^{3}+ 4x^{2} + 8x + 16) }{(x - 2)(x^{2} + 2x + 4)}$

$\sf \implies \lim_{x \to 2} \dfrac{(x^{4} + 2x^{3}+ 4x^{2} + 8x + 16) }{(x^{2} + 2x + 4)}$

Put the value of x = 2 in equation, we get.

$\sf \implies \lim_{x \to 2} \dfrac{((2)^{4} + 2(2)^{3}+ 4(2)^{2} + 8(2) + 16) }{((2)^{2} + 2(2) + 4)}$

$\sf \implies \lim_{x \to 2} \dfrac{(16 + 16+ 16 + 16 + 16) }{(4 + 4 + 4)}$

$\sf \implies \lim_{x \to 2} \dfrac{80}{12}$

$\sf \implies \lim_{x \to 2} = \dfrac{20}{3}$

                                                                                                                         

MORE INFORMATION.

Logarithmic inequality.

Let a is real number, such that.

(1) = For a > 1 the inequality ㏒ₐx > ㏒ₐy & x > y are equivalent.

(2) = If a > 1 then ㏒ₐx < α ⇒ 0 < x < $\sf a^{\alpha }$ .

(3) = If a > 1 then ㏒ₐx > α ⇒ x > $\sf a^{\alpha }$.

(4) = For 0 < a < 1 the inequality 0 < x < y & ㏒ₐx > ㏒ₐy are equivalent.

(5) = If 0 < a < 1 then ㏒ₐx < α ⇒ x > $\sf a^{\alpha }$.

Answer 2

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used \::}}}} \end{gathered}$

$\boxed{ \blue{\bf \:\lim_{x\to \: a}\dfrac{ {x}^{n} - {a}^{n} }{x - a} = {na}^{n - 1} }}$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\large\underline\purple{\bold{Solution :- }}$

$\bf \:\lim_{x\to2}\dfrac{ {x}^{5} - 32 }{ {x}^{3} - 8 }$

On substituting directly x = 0, we get indeterminant form

$\rm : \implies \:\:\lim_{x\to2}\dfrac{ {x}^{5} - {2}^{5} }{ {x}^{3} - {2}^{3} }$

Now, divide both terms by x - 2, we get

$\rm : \implies \:\:\lim_{x\to2}\dfrac{\dfrac{{x}^{5} - {2}^{5}}{x - 2} }{\dfrac{{x}^{3} - {2}^{3}}{x - 2} }$

$\rm : \implies \:\dfrac{5 {(2)}^{5 - 1} }{3 {(2) }^{3 - 1} }$

$\rm : \implies \:\dfrac{5 \times {(2)}^{4} }{3 \times {(2)}^{2} }$

$\rm : \implies \:\dfrac{5}{3} \times 4$

$\rm : \implies \:\dfrac{20}{3}$

$\boxed {\pink{\rm : \implies \:\:\lim_{x\to2}\dfrac{ {x}^{5} - 32}{ {x}^{3} - 8} = \dfrac{20}{3} }}$

─━─━─━─━─━─━─━─━─━─━─

Explore more ;-

$\begin{gathered}(1)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: \dfrac{sin \: x}{x} \: = \: 1 }}}}}} \\ \end{gathered}$

$\begin{gathered}(2)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: \dfrac{tan \: x}{x} \: = \: 1 }}}}}} \\ \end{gathered}$

$\begin{gathered}(3)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: \dfrac{ {e}^{x} - 1 }{x} \: = \: 1 }}}}}} \\ \end{gathered}$

$\begin{gathered}(4)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: \dfrac{ log(1 + x) }{x} \: = \: 1 }}}}}} \\ \end{gathered}$

$\begin{gathered}(5)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: \dfrac{ {a}^{x} - 1}{x} \: = \: log \: a }}}}}} \\ \end{gathered}$