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  • Using the method of L’ Hospital, evaluate the limit. <img alt="\lim _{x \rightarrow \infty} \frac{2 x^3-3 x^2-4 x+1}{x^3+2 x^2-5 x}" src="https://entrancecorner.

Using the method of L’ Hospital, evaluate the limit.

\lim _{x \rightarrow \infty} \frac{2 x^3-3 x^2-4 x+1}{x^3+2 x^2-5 x}

Option: 1

\infty


Option: 2

0


Option: 3

2


Option: 4

Cannot be determined


Answers (1)

best_answer

Apply the L' Hospital's Rule, when  \lim _{x \rightarrow 0} \frac{f(x)}{g(x)}=\frac{0}{0}, \frac{\infty}{\infty}, \text { etc. }  [an intermediate form] in the following way

  • Differentiate \lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\lim _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)}, provided that the limit \lim _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)} exists.

  • Otherwise, go on differentiating till the limit with the determinate form is achieved.

The provided limit is

\lim _{x \rightarrow \infty} \frac{2 x^3-3 x^2-4 x+1}{x^3+2 x^2-5 x}=\frac{\infty}{\infty}

Here use the L’ Hospital rule and also refer to the following formula.

\frac{d}{d x} x^n=n x^{n-1}

Now, derive the following:

\begin{aligned} & \lim _{x \rightarrow \infty} \frac{2 x^3-3 x^2-4 x+1}{x^3+2 x^2-5 x} \\ & =\lim _{x \rightarrow \infty} \frac{\frac{d}{d x}\left(2 x^3-3 x^2-4 x+1\right)}{\frac{d}{d x}\left(x^3+2 x^2-5 x\right)} \\ & =\lim _{x \rightarrow \infty} \frac{\left(2 \times 3 x^2-3 \times 2 x-4\right)}{\left(3 x^2+2 \times 2 x-5\right)} \quad\left[x \frac{d}{d x} x^n=n x^{n-1}\right] \\ & =\lim _{x \rightarrow \infty} \frac{\left(6 x^2-6 x-4\right)}{\left(3 x^2+4 x-5\right)} \quad\left[=\frac{\infty}{x}\right] \end{aligned}

\begin{aligned} & =\lim _{x \rightarrow \infty} \frac{d x}{d}\left(6 x^2-6 x-4\right) \\ & =\lim _{x \rightarrow \infty} \frac{(6 \times 2 x-6)}{(3 \times 2 x+4)} \quad\left[3 \frac{d}{d x} x^n=n x^{n-1}\right] \\ & =\lim _{x \rightarrow \infty} \frac{6(2 x-1)}{2(3 x+2)} \quad\left[=\frac{\infty}{x}\right] \end{aligned}

\begin{aligned} & =3 \lim _{x \rightarrow \infty} \frac{\frac{a}{d x}(2 x-1)}{\frac{d}{d x}(3 x+2)} \\ & =3 \lim _{x \rightarrow \infty} \frac{2}{3} \\ & =2 \end{aligned}

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