Question

Q1.Limit x tends to infinity
5x^2+3x+1/x^2-x+1
Answer is 5

Answer 1

Step-by-step explanation:

Solution :

$\begin{gathered}:\implies \sf \lim_{x \to \infty} \dfrac{5x^{2} + 3x + 1}{x^{2} - x + 1} \\ \\ \end{gathered} </p><p>$

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

$\begin{gathered}:\implies \sf \lim_{x \to \infty} \dfrac{5x^{2} + 3x + 1}{x^{2} - x + 1} = \lim_{x \to \infty} \dfrac{\dfrac{5x^{2} + 3x + 1}{x^{2}}}{\dfrac{x^{2} - x + 1}{x^{2}}} \\ \\ \end{gathered} </p><p>$

$\begin{gathered}:\implies \sf \lim_{x \to \infty} \dfrac{5x^{2} + 3x + 1}{x^{2} - x + 1} = \lim_{x \to \infty} \dfrac{\dfrac{5x^{2}}{x^{2}} + \dfrac{3x}{x^{2}} + \dfrac{1}{x^{2}}}{\dfrac{x^{2}}{x^{2}} - \dfrac{x}{x^{2}} + \dfrac{1}{x^{2}}} \\ \\ \end{gathered} </p><p>$

$\begin{gathered}:\implies \sf \lim_{x \to \infty} \dfrac{5x^{2} + 3x + 1}{x^{2} - x + 1} = \lim_{x \to \infty} \dfrac{5 + \dfrac{3}{x} + \dfrac{1}{x^{2}}}{1 - \dfrac{1}{x} + \dfrac{1}{x^{2}}} \\ \\ \end{gathered} </p><p>$

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

$\begin{gathered}:\implies \sf \lim_{x \to \infty} \dfrac{5x^{2} + 3x + 1}{x^{2} - x + 1} = \dfrac{5 + 0 + 0}{1 - 0 + 0} \\ \\ \end{gathered} </p><p>$

$\begin{gathered}:\implies \sf \lim_{x \to \infty} \dfrac{5x^{2} + 3x + 1}{x^{2} - x + 1} = 5 \\ \\ \end{gathered} </p><p>$

$\begin{gathered}\boxed{\therefore \sf \lim_{x \to \infty} \dfrac{5x^{2} + 3x + 1}{x^{2} - x + 1} = 5} \\ \\ \end{gathered} </p><p>$

Answer 2

Answer:

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