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Find Limit : \lim _{x \rightarrow 2} \frac{\left(3 x^2+2 x-5\right)}{\left(x^2-1\right)}

Option: 1

\frac{10}{3}


Option: 2

\frac{11}{3}


Option: 3

\frac{11}{5}


Option: 4

\frac{7}{5}


Answers (1)

best_answer

The expression's denominator is zero at x=2 an indeterminate form of the form (nonzero constant) / 0. Factor the numerator and denominator together to calculate the limit:

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

Canceling out the common factor of (x-1)

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

substitute x=2  into this expression to get:

\begin{aligned} & \frac{3(2)+5}{2+1} \\ & =\frac{11}{3} \end{aligned}

 

Posted by

Kuldeep Maurya

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