Question

If $f\left(x\right)=-2x^5+x^3-5x^2+7$, then $\underset{x\to -\infty }{\lim }f\left(x\right)=\_\_\_\_\_\_\_\_$and$\underset{x\to \infty }{\lim }f\left(x\right)=\_\_\_\_\_\_\_\_$.

Short answer

The solution is$\underset{x\to -\infty }{\lim }f\left(x\right)=\infty$and$\underset{x\to \infty }{\lim }f\left(x\right)=-\infty$.

Step-by-step solution

Given information 

Function:$f\left(x\right)=-2x^5+x^3-5x^2+7$

Calculation

We have,

$f\left(x\right)=-2x^5+x^3-5x^2+7$

Now, $\underset{x\to -\infty }{\lim }f\left(x\right)=\underset{x\to -\infty }{\lim }\left(-2x^5+x^3-5x^2+7\right)$

role="math" localid="1646902515040" $=\underset{x\to -\infty }{\lim }\left(x^5\right)\left(-2+\frac{1}{x^2}-\frac{5}{x^3}+\frac{7}{x^5}\right)$

role="math" localid="1646902576542" $=\underset{x\to -\infty }{\lim }\left(x^5\right)\left(-2+0-0+0\right)$

$=\infty$

and $\underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }\left(-2x^5+x^3-5x^2+7\right)$

$=\underset{x\to \infty }{\lim }\left(x^5\right)\left(-2+\frac{1}{x^2}-\frac{5}{x^3}+\frac{7}{x^5}\right)$

$=\underset{x\to \infty }{\lim }\left(x^5\right)\left(-2+0-0+0\right)$

$=-\infty$