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༒ 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

Answer:

$\rm ∞$

Step-by-step explanation:

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

By L'Hôpitals rule,

$\rm \: \lim\limits_{x \to \infty} 4x-\dfrac{5}{2}$

$\rm = \lim\limits_{x \to \infty} \bigg( 4x \bigg)-\dfrac{5}{2}$

$\rm = 4 \lim\limits_{x \to \infty} \bigg( x \bigg)-\dfrac{5}{2}$

$\rm = ∞$

Answer 2

Answer:

∞

Step-by-step explanation:

We are asked to evaluate the following limit:

$\lim\limits_{x \to \infty} \left\{\dfrac{4x^{2} - 5x + 7}{2x - 3} \right\}$

We can see that factorisation of numerator is not possible, so let's divide all the terms by x.

$\implies\lim\limits_{x \to \infty} \left\{\dfrac{ \frac{4x^{2}}{x} - \frac{5x}x + \frac7x}{ \frac{2x}x - \frac 3x} \right\}$

$\implies\lim\limits_{x \to \infty} \left\{\dfrac{ 4x - 5+ \frac7x}{2- \frac 3x} \right\}$

Now substitute the limits.

$\implies\dfrac{ 4( \infty) - 5+ \frac7 \infty}{2- \frac 3 \infty}$

$\implies\dfrac{ \infty- 5+ 0 }{2- 0}$

$\implies\dfrac{ \infty }{2}$

$\implies \infty$

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

$\rule{280}{1}$

We can also solve this problem by using L'Hôpital's rule according to which if we have any limits which results 0/0 or ∞/∞ form by direct substitution, then the following results holds true.

${ \lim \limits_{ x \to \infty} \left \{ \dfrac{ f(x)}{g(x)} \right \}= \lim \limits_{ x \to \infty}\left \{ \dfrac{f'(x)}{g'(x)}\right \}}$

Here, f'(x) and g'(x) are derivatives of functions f and g respectively.

By using the definition of L'Hôpital's rule, we have:

${\lim\limits_{x \to \infty} \left\{\dfrac{4x^{2} - 5x + 7}{2x - 3} \right\} = \lim\limits_{x \to \infty} \left\{\dfrac{ \frac{d}{dx}(4x^{2} - 5x + 7)}{ \frac{d}{dx}(2x - 3)} \right\}}$

Now we will be using the following rules of derivative.

• $\boxed{ \dfrac{d}{dx}(x^n) = n x^{n-1}}$
• $\boxed{\dfrac{d}{dx}(k) = 0}$

$\implies\lim\limits_{x \to \infty} \left\{\dfrac{8x - 5}{2} \right\}$

Now by substituting the limits, we get:

$\implies\dfrac{8( \infty ) - 5}{2}$

$\implies\dfrac{ \infty }{2}$

$\implies \infty$

Hence, the given limit is equal to ∞.