Is this a valid way to find the limit of $f(x)$?

Question

If $\lim_{x\to 2}\frac{f(x)-5}{x-2}=3$, then find $\lim_{x\to 2}f(x)$.

$\lim_{x\to 2}\frac{f(x)-5}{x-2}=3$

$\frac{\lim_{x\to 2}f(x)-\lim_{x\to 2}5}{\lim_{x\to 2}x-\lim_{x\to 2}2}=3$

$\lim_{x\to 2}f(x)=3(\lim_{x\to 2}x-\lim_{x\to 2}2)+\lim_{x\to 2}5=5$

However, the textbook uses another method by manipulating $\lim_{x\to 2}f(x)-5$ to make it equal to $0$. Is my method okay or should I follow their method?

Answer 1

Suposse that $\lim\limits_{x\to 2} \dfrac{f(x)-5}{x-2}$ exists. We know that $\lim\limits_{x\to 2} x-2$ exists (moreover, is equal to $0$), then, the product of the limits exists and is equal to $0$ because one of the limits is zero, i.e. $$\lim\limits_{x\to 2} \dfrac{f(x)-5}{x-2}(x-2)=\lim\limits_{x\to 2} f(x)-5=0$$Now, we know that $\lim\limits_{x\to 2} 5$ exists (in fact, is equal to 5), thus, the sum of the limits exists and is equal to 5 because one limit is zero and the other is constant:$$\lim\limits_{x\to 2} f(x)-5+5=\lim\limits_{x\to 2} f(x)=5$$

Answer 2

Your method takes $\frac00=3$ and then cross multiplies. You can't cross multiply $\frac00$.

The method is text book refers to the point that if the limit of denominator is $0$ then the limit of the numerator must be $0$ for the limit to exist.

Answer 3

The question is about L'Hôpital's rule. Since the denominator goes to $0$ and the limit exists, then the numerator also goes to $0$, which means

$\displaystyle\lim_{x\to2}f(x)=5$