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  • Using the method of L' Hospital, evaluate the limit. <img alt="\lim _{x \rightarrow \infty} \frac{2 x^{2}+3 x+1}{3 x^{2}+5}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Clim%20_

Using the method of L' Hospital, evaluate the limit. \lim _{x \rightarrow \infty} \frac{2 x^{2}+3 x+1}{3 x^{2}+5}

Option: 1

\frac{2}{3}


Option: 2

\pm \frac{2}{3}


Option: 3

\frac{1}{3}


Option: 4

\frac{1}{2}


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

 

Posted by

Ritika Jonwal

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