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  • Evaluate the limit : <img alt="\lim _{x \rightarrow-2} \frac{2 x

Evaluate the limit : \lim _{x \rightarrow-2} \frac{2 x^3-3 x^2-4 x+1}{x^3+2 x^2-5 x}

Option: 1

\frac{1}{3}


Option: 2

\frac{2}{3}


Option: 3

\frac{19}{10}


Option: 4

Doesn’t exist


Answers (1)

best_answer

\begin{aligned} & \lim _{x \rightarrow-2} \frac{2 x^3-3 x^2-4 x+1}{x^3+2 x^2-5 x} \\ & =\lim _{x \rightarrow-2} \frac{x^3\left(2-\frac{3}{x}-\frac{4}{x^2}+\frac{1}{x^3}\right)}{x^3\left(1+\frac{2}{x}-\frac{5}{x^2}\right)} \\ & =\lim _{x \rightarrow-2} \frac{\left(2-\frac{3}{x}-\frac{4}{x^2}+\frac{1}{x^3}\right)}{\left(1+\frac{2}{x}-\frac{5}{x^2}\right)} \end{aligned}

\begin{aligned} & =\frac{2-\frac{3}{-2}-\frac{4}{(-2)^2}+\frac{1}{(-2)^3}}{1+\frac{2}{-2}-\frac{5}{(-2)^2}} \\ & =\frac{2+\frac{3}{2}-\frac{4}{4}-\frac{1}{8}}{1-1+\frac{5}{4}} \\ & =\frac{2+\frac{3}{2}-1-\frac{1}{8}}{1-1+\frac{5}{4}} \\ & =\frac{\frac{19}{8}}{\frac{5}{4}} \end{aligned}

=\frac{19}{10}

Posted by

Kuldeep Maurya

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