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  • Find limit:  <img alt="\lim _{x \rightarrow 3} \frac{\left(x^2-5 x+6\right)}{(x-3)^3}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Clim%20_%7Bx%20%5Crightarrow%203%7D%20%5Cfrac

Find limit:  \lim _{x \rightarrow 3} \frac{\left(x^2-5 x+6\right)}{(x-3)^3}

Option: 1

\frac{1}{6}


Option: 2

\frac{1}{4}


Option: 3

\frac{1}{7}


Option: 4

\frac{1}{5}


Answers (1)

As x\rightarrow 1 the expression's denominator changes to 0  and the numerator evaluates to (32-5(3)+6)=0.Consequently, we have a type 0 / 0 indeterminate form.by factoring the numerator:

\begin{aligned} & \frac{\left(x^2-5 x+6\right)}{(x-3)^3} \\ & =\frac{(x-3)(x-2)}{(x-3)^3} \end{aligned}

Simplifying this expression

\begin{aligned} & \frac{(x-3)(x-2)}{(x-3)^3} \\ & =\frac{(x-2)}{(x-3)^2} \end{aligned}

evaluate the limit using algebraic manipulation


\lim _{x \rightarrow 3} \frac{\left(x^2-5 x+6\right)}{(x-3)^3}

\begin{aligned} & =\lim _{x \rightarrow 3} \frac{(x-2)}{(x-3)^2} \\ & =\frac{(3-2)}{(3-3)^2} \\ & =\frac{1}{0} \end{aligned}

This is a type \infty / 0 indeterminate form. L'Hopital's rule, which stipulates that we take the derivative of the numerator and the denominator and reevaluate the limit if the limit of a quotient of functions is an indeterminate form of type 0 / 0 or \infty / \infty can be used to assess the limit.

\begin{aligned} & \lim _{x \rightarrow 3} \frac{\left(x^2-5 x+6\right)}{(x-3)^3} \\ & =\lim _{x \rightarrow 3} \frac{(x-2)}{(x-3)^2} \\ & =\lim _{x \rightarrow 3} \frac{1}{(2(x-3))} \\ & =\frac{1}{6} \end{aligned}

 

 

Posted by

Kshitij

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