Question

Evaluate lim
x->0 X² +5X/ X²+7x
​

Answer 1

EXPLANATION.

$\sf \implies \lim_{x \to 0} x^{2} + 5x/x^{2} + 7x.$

Put the value of x = 0 in equation.

$\sf \implies \lim_{x \to 0} \dfrac{x^{2} + 5x}{x^{2} + 7x.}$

$\sf \implies \lim_{x \to 0} \dfrac{(0)^{2} + 5(0)}{(0)^{2} + 7(0)} .$

As we can see that it is in the form of 0/0 form.

According to 0/0 form,

we can factorizes the equation, we get.

$\sf \implies \lim_{x \to 0} \dfrac{x(x + 5)}{x(x + 7)}.$

Again put the value of x = 0 in equation, we get.

$\sf \implies \lim_{x \to 0} \dfrac{(0 + 5)}{(0 + 7)}.$

$\sf \implies \lim_{x \to 0} = \dfrac{5}{7}.$

$\sf value \ of\ equation = \lim_{x \to 0} \dfrac{(x^{2} + 5x)}{(x^{2} + 7x)} = 5/7.$

                                                                                           

MORE INFORMATION.

By using some standard expansions,

(1) = Sin(x) = x - x³/3! + x⁵/5! -...

(2) = Cos(x) = 1 - x²/2! + x⁴/4! -..

(3) = Tan(x) = x + x³/3 + 2x⁵/15 +...

(4) = eˣ = 1 + x + x²/2! + x³/3! +...

(5) = e⁻ˣ = 1 - x + x²/2! - x³/3! +...  

Answer 2

Step-by-step explanation:

Given:-

$\bf : \implies {\lim_{x \to 0}\ x^2\ +\ \dfrac{5x}{x^2}\ +\ 7x.}$

$\frak{\underline{\underline{\dag By\ substituting\ the\ value\ of\ x\ =\ 0\ in\ equation,\ we\ get:-}}}$

$\bf : \implies {\lim_{x \to 0}\ \dfrac{x^2\ +\ 5x}{x^2\ +\ 7x}} \\ \\ \bf : \implies {\lim_{x \to 0}\ \dfrac{(0)^2\ +\ 5(0)}{(0)^2\ +\ 7(0)}}$

$\frak{\underline{\underline{\dag According\ to\ \dfrac{0}{0}\ form:-}}}$

$\bf : \implies {\lim_{x \to 0}\ \dfrac{x\ (x\ +\ 5)}{x\ (x\ +\ 7)}}$

$\frak{\underline{\underline{\dag By\ substituting\ the\ value\ of\ x\ =\ 0\ in\ equation,\ we\ get:-}}}$

$\bf : \implies {\lim_{x \to 0}\ \dfrac{(0\ +\ 5)}{(0\ +\ 7)}} \\ \\ \bf : \implies {\lim_{x \to 0}\ =\ \dfrac{5}{7}}$

$\frak{\underline{\underline{\dag Value\ of\ equation:-}}}$

$\bf : \implies {\lim_{x \to 0}\ \dfrac{(x^2\ +\ 5x)}{(x^2\ +\ 7x)}} \\ \\ \bf : \implies {\purple{\underline{\boxed{\bf \dfrac{5}{7}}}}}\bigstar$

Hence verified ✔.