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  • Using the method of L' Hospital, evaluate the limit. <img alt="\lim _{x \rightarrow 3} \frac{\left(x^{2}-4 x+3\right)}{\left(x^{2}-5 x+6\right)}" src="https://entrancecorner.oncodecogs.com

Using the method of L' Hospital, evaluate the limit. \lim _{x \rightarrow 3} \frac{\left(x^{2}-4 x+3\right)}{\left(x^{2}-5 x+6\right)}

Option: 1

1


Option: 2

2


Option: 3

\infty


Option: 4

0


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 3} \frac{\left(x^{2}-4 x+3\right)}{\left(x^{2}-5 x+6\right)}=\frac{0}{0}

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

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jitender.kumar

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