Question

lim x → ∞ 4x^2-5x+1/5x^2+2x+3​

Answer 1

Answer:

$\large \bold\red{\frac{4}{5} }$

Step-by-step explanation:

Given,

$\displaystyle\lim_{x \to \infty} \frac{4 {x}^{2} - 5x + 1}{5 {x}^{2} + 2x + 3 }$

Clearly,

It's ∞/∞ form, when x = ∞.

Therefore,

We can apply L Hospital's Rule here.

Therefore,

We get,

$= \displaystyle\lim_{x \to \infty} \frac{ \frac{d}{dx} (4 {x}^{2} - 5x + 1)}{ \frac{d}{dx}(5 {x}^{2} + 2x + 3) } \\ \\ = \displaystyle\lim_{x \to \infty} \frac{8x - 5}{10x + 2}$

Again,

It's in ∞/∞ form.

Therefore,

Applying L Hospital's Rule,

We get,

$= \displaystyle\lim_{x \to \infty} \frac{ \frac{d}{dx}(8x - 5) }{ \frac{d}{dx}(10x + 2) } \\ \\ = \displaystyle\lim_{x \to \infty} \frac{8}{10} \\ \\ = \large \bold{\frac{4}{5} }$

Answer 2

$\LARGE\underline{\underline{\sf \red{T}\blue{o}\:\green{F}\orange{i}\pink{n}\red{d}:}}$

correct Question

lim x → ∞ 4x^2-5x+1/5x^2+2x-3

$\LARGE\underline{\underline{\sf \red{S}\blue{o}\green{l}\orange{u}\pink{t}\purple{i}\orange{o}\red{n}:}}$

Lets solve first numerator and denominator d/dx first by

d/dx (x^n+Bx+A) = nx+B

in numerator we have

$\frac{d}{dx} (4 {x}^{2} - 5x + 1) \\ \\ \frac{d}{dx} (2 \times 4x - 5) \\ \\ (8x - 5)$

in denominator we have ,

$\frac{d}{dx} (5 {x}^{2} + 2x - 3 )\\ \\ \frac{d}{dx} (2 \times 5x + 2) \\ \\ (10x + 2)$

Now we have ,

$lim x → ∞( \frac{8x - 5}{10x + 2} )$

Its now again in ∞/∞ form,

again Differentiation , it we get,

As we know d/dx (nx+b) = n ........

So,

$lim x → ∞ \: \frac{ \frac{d}{dx}(8x - 5) }{ \frac{d}{dx}(10x + 2) } \\ \\ lim x → ∞ \: \: \frac{8}{10}$

$\huge{\frac{4}{5}}$

$\huge\underline\mathfrak\green{Hope\:it\:Helps\:You}$