MHB How can we calculate the limits?

[mathmari]

Hey!

Could you give me a hint how we could calculate the following limits?

$\displaystyle{\lim_{x\rightarrow -1^-}\frac{-x}{\sqrt{1-x^2}}}$

$\displaystyle{\lim_{x\rightarrow -1^+}\frac{-x}{\sqrt{1-x^2}}}$ $\displaystyle{\lim_{x\rightarrow 1^-}\frac{-x}{\sqrt{1-x^2}}}$

$\displaystyle{\lim_{x\rightarrow 1^+}\frac{-x}{\sqrt{1-x^2}}}$

(Wondering)

 

[I like Serena]

Hey mathmari! (Smile)

When $x>1$, the expression $\frac{-x}{\sqrt{1-x^2}}$ is undefined.
Therefore:
$$\lim_{x\rightarrow 1^+}\frac{-x}{\sqrt{1-x^2}} \text{ does not exist}$$

And when $x<1$ and $x\to 1^-$, the numerator approaches $-1$, while the denominator approaches $0^+$, so that $\frac{-x}{\sqrt{1-x^2}} \to -\infty$
Thus:
$$\lim_{x\rightarrow 1^-}\frac{-x}{\sqrt{1-x^2}} \text{ does not exist}$$

 

[mathmari]

When $x>1$, the expression $\frac{-x}{\sqrt{1-x^2}}$ is undefined.
Therefore:
$$\lim_{x\rightarrow 1^+}\frac{-x}{\sqrt{1-x^2}} \text{ does not exist}$$

And when $x<1$ and $x\to 1^-$, the numerator approaches $-1$, while the denominator approaches $0^+$, so that $\frac{-x}{\sqrt{1-x^2}} \to -\infty$
Thus:
$$\lim_{x\rightarrow 1^-}\frac{-x}{\sqrt{1-x^2}} \text{ does not exist}$$

I got it! Thank you very much! (Happy)