Showing that $\lim_{x \to -\infty}(x+{1\over x}) = -\infty$

Question

My attempt. I' not sure about this, but I think the definition for this limit will be:

$\forall M <0 \space \exists N < 0, \space x < N \Rightarrow f(x) < M$

We start off knowing that $x < N$ $\Rightarrow$ $x^2 > N^2\Rightarrow {1\over x^2 } + 1 < {1 \over N^2} + 1 \Rightarrow x({1\over x^2 } + 1) < N({1 \over N^2} + 1)$. Now, since ${1\over N^2} + 1 > 1$ and $N < 0$, we have that $N({1\over N^2} + 1 ) < N$. Thefore, $x({1\over x^2} + 1) < N$, which mean that we will choose N = M in our proof.

Proof:

Pick N = M. Since $x < N \Rightarrow x({1\over x^2 } + 1) < N({1 \over N^2} + 1)< N = M$, as required.

Is this correct?

Answer 1

Much to complicated. You have that $\frac{1}{x}\leq 0$ when $x<0$. Let $M<0$. Set $N=M$. Therefore, if $x<N$, then $$x+\frac{1}{x}<M,$$ what prove the claim.

Answer 2

Note that by squeeze theorem

$$x+{1\over x}<x\to-\infty$$

Answer 3

Try using the fact that $$x > (x + 1/x)$$ when $x<0$. The proof is then immediate.

Answer 4

You wanted to evaluate the limit from first principles, so

Yes, you want to prove :

$\forall M <0,\space \exists N < 0, \space x < N \Rightarrow f(x) < M$

"We start off knowing that $x < N$ $\Rightarrow$ $x^2 > N^2\Rightarrow {1\over x^2 } + 1 < {1 \over N^2} + 1$" This is correct.

$\Rightarrow x({1\over x^2 } + 1) < N({1 \over N^2} + 1)$.

This is wrong! Consider $x=-0.5,N=-0.4,x<N$, but

$\Rightarrow x({1\over x^2 } + 1)=-2.5 > N({1 \over N^2} + 1)=-2.9!$

You are making an invalid comparison here. Instead you can simply do this:

Proof:

Pick N = M. Since $x < N,x^2>xM$,

$x^2+1>x^2>xM$, since obviously $x\neq 0$,

$f(x)=x+\frac{1}{x}<x<M$, as required.