Question

Solve the limit :-

$\large\lim \limits_ { x \rightarrow \infty } \left(\frac { 4 x ^ { 2 } - 5 x + 7 } { 2 x - 3 } \right )$

Is the answer undefined or infinity?
I want full solution ​

Answer 1

Answer:

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$\sf \! \lim\limits_{x \to \infty} \bigg\{\dfrac{4x^{2} - 5x + 7}{2x - 3} \bigg\} = \infty$

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Step-by-step explanation:

We've been asked to solve the following limit:

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$\Longrightarrow \sf \! \lim\limits_{x \to \infty} \bigg\{\dfrac{4x^{2} - 5x + 7}{2x - 3} \bigg\}$

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Here, 'x' approaches/tends to ∞.

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The degree (highest power) of the rational function is 2.

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The fraction is numerator-heavy as squaring a very large number and multiplying it by a positive number (4x²; x ➝ ∞) would result in an enormous (+ve) number. This makes the other terms in the numerator that are being added and subtracted to 4x² insignificant as they're very small numbers when compared to 4 times the square of a very large number.

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Right now by looking at the fraction, we cannot accurately determine what both the numerator and denominator approach. So let's try to reduce the fraction.

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In order to do so, we can divide the numerator and the denominator by the highest power of the denominator. Or, we can factor out the highest power out of both the numerator and denominator. Doing both and simplifying will give us the same result, I'm following the first method for this solution. The highest power in the denominator here is x¹.

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$\Longrightarrow \sf \! \lim\limits_{x \to \infty} \bigg\{\dfrac{(4x^{2}/x) - (5x/x) + (7/x)}{(2x/x) - (3/x)} \bigg\}$

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$\Longrightarrow \sf \! \lim\limits_{x \to \infty} \bigg\{\dfrac{4x - 5 + (7/x)}{2 - (3/x)} \bigg\}$

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We know that,

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$\boxed{\sf \lim_{x \to n} \bigg\{ \frac{f(x)}{g(x)} \bigg\} = \dfrac{\lim \limits_{x \to n} \Big\{ f(x) \Big\}}{\lim\limits_{x \to n} \Big\{ g(x) \Big\}} \ \ \bigg[Where \ \lim\limits_{x \to n} g(x) \neq 0)\bigg]}$

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$\Longrightarrow \sf \! \ \dfrac{\lim\limits_{x \to \infty} \Big\{4x - 5 + (7/x)\Big\}}{\lim\limits_{x \to \infty}\Big\{2 - (3/x)\Big\}}$

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Finding lim(x ➝ ∞){4x - 5 + (7/x)}:

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$\Longrightarrow \sf \! \lim\limits_{x \to \infty} \bigg\{4x - 5 + \dfrac{7}{x} \bigg\}$

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$\Longrightarrow \sf \! \lim\limits_{x \to \infty} \bigg\{4x\bigg\} - \lim\limits_{x \to \infty}\bigg\{5\bigg\} + \lim\limits_{x \to \infty}\bigg\{\dfrac{7}{x} \bigg\}$

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$\Longrightarrow \sf 4 \times \lim\limits_{x \to \infty}\bigg\{x\bigg\} - \lim\limits_{x \to \infty}\bigg\{5\bigg\} + \lim\limits_{x \to \infty}\bigg\{\dfrac{7}{x} \bigg\}$

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On applying the following rules . . .

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$\sf \longrightarrow \ \lim\limits_{x \to a}\Big\{Constant\Big\} = Constant$

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$\sf \longrightarrow \lim\limits_{x \to \infty}\bigg\{\dfrac{N}{x^{r}} \bigg\} = 0 : \textsf{[Where N = A natural number, r = +ve rational number]}$

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$\sf \longrightarrow \ \lim\limits_{x \to \infty}\Big\{x\Big\} = \infty$

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. . . we get,

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$\Longrightarrow \sf \ (4 \times \infty) - 5 + 0$

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$\Longrightarrow \sf \infty - 5$

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$\Longrightarrow \sf \infty$

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Similarly, finding lim(x ➝ ∞){2 - (3/x)}:

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$\Longrightarrow \sf \! \lim\limits_{x \to \infty} \bigg\{2 - \dfrac{3}{x} \bigg\}$

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$\Longrightarrow \sf \! \lim\limits_{x \to \infty} \bigg\{2\bigg\} - \lim\limits_{x \to \infty} \bigg\{\dfrac{3}{x} \bigg\}$

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$\Longrightarrow \sf 2 - 0$

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$\Longrightarrow \sf 2$

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Substituting these values in our original limit we get,

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$\Longrightarrow \sf \! \dfrac{\lim\limits_{x \to \infty} \Big\{4x - 5 + (7/x)\Big\}}{\lim\limits_{x \to \infty}\Big\{2 - (3/x)\Big\}}$

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$\Longrightarrow \sf \! \dfrac{\infty}{2}$

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We know that infinity divided by a finite number gives infinity.

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$\Longrightarrow \sf \infty$

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Therefore,

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$\Longrightarrow \sf \! \lim\limits_{x \to \infty} \bigg\{\dfrac{4x^{2} - 5x + 7}{2x - 3} \bigg\} = \infty$

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Hence solved.

Answer 2

Answer:

$\underline{\purple{\ddot{\Maths dude}}}$

Given:-

• limit x tends to infinity (4x^2-5x+7)/2x-3.

To prove :-

• Solving the limit and we should find the answer as undefined or infinity.

Explanation :-

• You can refer the attachment for answer

Hope it helps u @Pro Thala

Thank you .