MHB Limit as x goes to 5 from below

[Guest2]

Dividing by the highest power for $\displaystyle \lim_{x \to 5^{-}}\frac{x^{100}-4x^{99}}{x-5}$ I get

$\displaystyle \lim_{x \to 5^{-}}\frac{x^{100}-4x^{99}}{x-5}= \lim_{x \to 5^{-}}\frac{1-4/x}{1/x^{99}-5/x^{100}}$

However the denominator goes to $0$ whereas the numerator goes to $1-4/5$

Why isn't dividing by the highest power working?

 

[kaliprasad]

Dividing by the highest power for $\displaystyle \lim_{x \to 5^{-}}\frac{x^{100}-4x^{99}}{x-5}$ I get

$\displaystyle \lim_{x \to 5^{-}}\frac{x^{100}-4x^{99}}{x-5}= \lim_{x \to 5^{-}}\frac{1-4/x}{1/x^{99}-5/x^{100}}$

However the denominator goes to $0$ whereas the numerator goes to $1-4/5$

Why isn't dividing by the highest power working?

the reason in the given expression numerator is not zero but denominator is zero at x = 5.
hence after division also

 

[Guest2]

the reason in the given expression numerator is not zero but denominator is zero at x = 5.
hence after division also

So how does one calculate the limit?

 

[kaliprasad]

So how does one calculate the limit?

\
so the limit does not exist as it goes to infinite

 

[Prove It]

Kali, you are correct that the limit does not exist, but the LEFT HAND limit can still be evaluated.

To the OP - what happens when you divide by a very, very small number (close to 0)?

 

[kaliprasad]

Kali, you are correct that the limit does not exist, but the LEFT HAND limit can still be evaluated.

To the OP - what happens when you divide by a very, very small number (close to 0)?

you mean numerator is is positive and denominator is -ve( close to zero) so limit is - infinite ?

 

[Prove It]

you mean numerator is is positive and denominator is -ve( close to zero) so limit is - infinite ?

That is correct :)