Question

༒ Bʀᴀɪɴʟʏ Sᴛᴀʀs

༒Mᴏᴅᴇʀᴀᴛᴇʀs

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Solve the limit :-

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


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Answer 1

$\large\underline{\text{Main idea}}$

$\cdots\longrightarrow\boxed{\displaystyle\lim_{x\to\infty}\dfrac{1}{x}=0}$

$\large\underline{\text{Explanation}}$

The limit will diverge to the positive infinity, as the degree of the numerator is greater than that of the denominator.

Let's check some properties of limits to solve the problem.

$\boxed{\begin{aligned}\displaystyle&(1)\lim_{x\to a}kf(x)=k\lim_{x\to a}f(x)\\\\&(2)\lim_{x\to a}f(x)g(x)=\lim_{x\to a}f(x)\lim_{x\to a}g(x)\end{aligned}}$

The calculation of the given limit is as follows.

$\boxed{\begin{aligned}\displaystyle\lim_{x\to\infty}\dfrac{4x^{2}-5x+7}{2x-3}&=\lim_{x\to\infty}\dfrac{\dfrac{1}{x}(4x^{2}-5x+7)}{\dfrac{1}{x}(2x-3)}\\\\&=\lim_{x\to\infty}\dfrac{4x-5+\dfrac{7}{x}}{2-\dfrac{3}{x}}\ \text{[ \because Property (2)]}\\\\&=\lim_{x\to\infty}\dfrac{4x-5}{2}\\\\&=2\times \infty-\dfrac{5}{2}\ \text{[ \because Property (1)]}\\\\&=\infty\end{aligned}}$

The limit diverges to the positive infinity.

$\large\underline{\text{Extra information}}$

In the same manner that we derived the limit of a rational function $\displaystyle\lim_{x\to\infty}\dfrac{f(x)}{g(x)}$,

(1) the limit diverges if and only if the degree of $f(x)$ is greater than that of $g(x)$.

(2) the limit converges to the fraction of the highest coefficients, if and only if the degree of $f(x)$ is greater than that of $g(x)$.

(3) the limit converges to zero if and only if the degree of $f(x)$ is less than that of $g(x)$.