Finding a limit of a rational function when the function isnt moving towards 0

Question

Consider this:

$$\lim_{x \to 2} \frac{x^2+3x-4}{x^2-7x+10}$$ Now I've solved this for the case when x is approaching 0,but when I try to use the same method for when it is approaching 2,I dont get very far.So first I put in 2 and got that in the numerator we get a 2 and denominator we get a 0,so division by 0 we want to avoid that.Now I need to "change" the look of my polynomials to get the limit,but I'm not sure how to do that? I am not sure if I can use L'Hospital here since it isn't 0/0 but rather 2 divided by zero? A bit of advice would be much appreciated

Many thanks!

Answer 1

Note that $x^2-7x+10=(x-2)(x-5)$. Therefore your limit is$$\lim_{x\to2}\frac{\frac{x^2+3x-4}{x-5}}{x-2}.\tag1$$Since $\lim_{x\to2}\frac{x^2+3x-4}{x-5}=-2$, the limit $(1)$ doesn't exist, because$$\lim_{x\to2^+}\frac{\frac{x^2+3x-4}{x-5}}{x-2}=-\infty\quad\text{and}\quad\lim_{x\to2^-}\frac{\frac{x^2+3x-4}{x-5}}{x-2}=\infty.$$

Answer 2

The rational function can be written as $$ q(x)=\frac{x^2+3x-4}{x^2-7x+10}=\frac{(x+4)(x-1)}{(x-2)(x-5)}=\frac{1}{x-2}\cdot \frac{(x+4)(x-1)}{x-5}=:\frac{1}{x-2}\cdot g(x) $$

Note that $$ \lim_{x\to 2}g(x)=-2 $$ But $\displaystyle \lim_{x\to 2}\frac{1}{x-2}$ does not exist.^$\dagger$ So the whole limit $\lim_{x\to 2}q(x)$, does not exist since otherwise, by one of the limit laws: $$ \lim_{x\to 2}\frac{1}{x-2}=\frac{\lim_{x\to 2}q(x)}{\lim_{x\to 2}g(x)} $$

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^$\dagger$ Notes.

This is a case where you should go over the materials about two-sided limits.

The following two limits show that $\displaystyle \lim_{x\to 0}\frac{1}{x}$ does not exist: $$ \lim_{x\to 0-}\frac{1}{x},\quad \lim_{x\to 0+}\frac{1}{x} $$

In general, $\lim_{x\to a}f(x)$ exists if and only if $$ \lim_{x\to a-}f(x)=\lim_{x\to a+}f(x)=L $$ where $L$ is a real number. $L$ could also be $+\infty$ or $-\infty$ if you want to deal with generalized limits.