Finding a limit at negative infinity

Question

$$\lim_{x \to -\infty}{x^2 + 2 \cdot x^7}$$

Hello,

I know that the answer to this is $- \infty$, but I'm having trouble showing it. Can anyone please point me in the right direction?

Thanks!

Answer 1

Let $x = -\frac{1}{t}$, now as $$\lim_{x \to -\infty}{x^2 + 2 \cdot x^7}$$ we can say, as $x$ tends to $-\infty$, $t$ would tend to $0$. Note the negative sign. Change the problem as: $$\lim_{t \to 0}{\frac{1}{t^2} - \frac{2}{t^7}}$$ Solve to get: $$\lim_{t \to 0}{\frac{t^5-2}{t^7}} = -\frac{2}{0} = -\infty$$ Hope this helps...

Answer 2

You can solve it as follows:

$$\lim_{x \to -\infty}{x^2 + 2 \cdot x^7}$$ $$=\lim_{x \to -\infty}{x^7(\frac{1}{x^5} + 2)}$$ $$=-\infty(\frac{1}{-\infty} + 2)$$ $$=-\infty(0 + 2)$$ $$=-\infty$$

Answer 3

As $x\to -\infty$ we have $$3x^7\le x^2+2x^7\le x^7$$ now squeeze....

Answer 4

$$\lim_{x \to -\infty}(x^2 + 2 x^7)=-\infty$$ For any $M>0$ there exists an $N>0$ such that if $x<-N$ then $f(x)<-M$

That is $x^2 + 2 x^7<-M$.

For $x<0$ we have $2x^7<x^2+2x^7<-M$ therefore taking $N=\sqrt[7]{\frac{M}{2}}$ if $x<-N$ we have $f(x)<-M$