Question

Limit x tends to infinity
5x^2+3x+1/x^2-x+1
Answer is 5

Answer 1

Solution :

$:\implies \sf \lim_{x \to \infty} \dfrac{5x^{2} + 3x + 1}{x^{2} - x + 1}$

By dividing x²(Here, the greatest power of x) to both the Numerator and Denominator in the equation, we get :-

$:\implies \sf \lim_{x \to \infty} \dfrac{5x^{2} + 3x + 1}{x^{2} - x + 1} = \lim_{x \to \infty} \dfrac{\dfrac{5x^{2} + 3x + 1}{x^{2}}}{\dfrac{x^{2} - x + 1}{x^{2}}}$

$:\implies \sf \lim_{x \to \infty} \dfrac{5x^{2} + 3x + 1}{x^{2} - x + 1} = \lim_{x \to \infty} \dfrac{\dfrac{5x^{2}}{x^{2}} + \dfrac{3x}{x^{2}} + \dfrac{1}{x^{2}}}{\dfrac{x^{2}}{x^{2}} - \dfrac{x}{x^{2}} + \dfrac{1}{x^{2}}}$

$:\implies \sf \lim_{x \to \infty} \dfrac{5x^{2} + 3x + 1}{x^{2} - x + 1} = \lim_{x \to \infty} \dfrac{5 + \dfrac{3}{x} + \dfrac{1}{x^{2}}}{1 - \dfrac{1}{x} + \dfrac{1}{x^{2}}}$

We know that, when limit x → ∞, then 1/x = 0 (Same 1/x² will be 0), so by applying it here, we get :

$:\implies \sf \lim_{x \to \infty} \dfrac{5x^{2} + 3x + 1}{x^{2} - x + 1} = \dfrac{5 + 0 + 0}{1 - 0 + 0}$

$:\implies \sf \lim_{x \to \infty} \dfrac{5x^{2} + 3x + 1}{x^{2} - x + 1} = 5$

$\boxed{\therefore \sf \lim_{x \to \infty} \dfrac{5x^{2} + 3x + 1}{x^{2} - x + 1} = 5}$

Answer 2

Answer:

$\lim_{x \rightarrow \infty}$ => 5

Step-by-step explanation:

$\lim_{x \rightarrow \infty}$ = $\dfrac{5x^2+3x+1}{x^2-x+1}$

Dividing x^2 in all the numbers at denominator:

$\lim_{x \rightarrow \infty}$ = $\sf{(5x^2/x^2)+(3x/x^2)+(1/x^2) /(x^2/x^2)-(x/x^2)+(1/x^2)}$

$\lim_{x \rightarrow \infty}$ = $\sf{(5+(3/x)+(1/x^2) /1-(1/x)+(1/x^2)}$

$\lim{x \rightarrow \infty}$ => 1/x

So, $\lim{x \rightarrow \infty}$ is also => $\dfrac{1}{x^2}$ = 0

So, $\lim_{x \rightarrow \infty}$ = $\sf{(5+(3/x)+(1/x^2) /1-(1/x)+(1/x^2)}$ = $\dfrac{5+ 3(0)+0}{(1-0+0)}$

$\dfrac{5+ 0+0}{(1-0+0)}$

= 5

Hence,

Final answer:

$\lim_{x \rightarrow \infty}$ => 5