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  • Using the method of L’ Hospital, evaluate the limit. <img alt="\lim _{x \rightarrow \infty} \frac{x(4 x+1)^2}{(x+4)\left(x^2+6 x-1\right)}" src="https://entrancecorner.oncodecogs.co

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

\lim _{x \rightarrow \infty} \frac{x(4 x+1)^2}{(x+4)\left(x^2+6 x-1\right)}

Option: 1

16


Option: 2

4


Option: 3

24


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

\begin{aligned} & \lim _{x \rightarrow \infty} \frac{x(4 x+1)^2}{(x+4)\left(x^2+6 x-1\right)} \\ & =\lim _{x \rightarrow \infty} \frac{\left(16 x^3+8 x^2+x\right)}{\left(x^3+10 x^2+23 x-4\right)} \\ & =\frac{\infty}{\infty} \end{aligned}

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{\left(16 x^3+8 x^2+x\right)}{\left(x^3+10 x^2+23 x-4\right)} \\ & =\lim _{x \rightarrow \infty} \frac{\frac{d}{d x}\left(16 x^3+8 x^2+x\right)}{\frac{d}{d x}\left(x^3+10 x^2+23 x-4\right)} \\ & =\lim _{x \rightarrow \infty} \frac{\left(16 \times 3 \times x^2+8 \times 2 \times x\right)}{\left(3 \times x^2+10 \times 2 \times x+23\right)} \quad\left[ \frac{d}{d x} x^n=n x^{n-1}\right] \\ & =\lim _{x \rightarrow \infty} \frac{16\left(3 x^2+x\right)}{\left(3 x^2+20 x+23\right)} \quad\left[=\frac{\infty}{\infty}\right] \\ & =\lim _{x \rightarrow \infty} \frac{16}{\frac{d}{d x}\left(3 x^2+20 x+23\right)} \end{aligned}

\begin{aligned} & =\lim _{x \rightarrow+\infty} \frac{16(3 \times 2 \times x+1)}{(3 \times 2 \times x+20)} \\ & =\lim _{x \rightarrow \infty} \frac{16(6 x+1)}{(6 x+20)} \quad\left[=\frac{\infty}{\infty}\right] \\ & =\lim _{x \rightarrow \infty} \frac{16 \frac{d}{d x}(6 x+1)}{\frac{d}{d x}(6 x+20)} \\ & =\lim _{x \rightarrow \infty} \frac{16 \times 6}{6} \\ & =16 \end{aligned}

Posted by

Rakesh

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