Negative limit answer
Find $\lim_{x\to 13} {5x-65 \over 169-x^2}.$
My answer is $5 \over 26$.
Symbolab and WolfRam answer $-{5 \over 26}$.
Where does the negative come from?
$\endgroup$3 Answers
$\begingroup$$\dfrac{5x-65}{169-x^2}=\dfrac{-5(13-x)}{(13-x)(13+x)}=\dfrac{-5}{13+x}$
$\endgroup$ $\begingroup$You have $(x-13)$ at numerator and $(13-x)$ at denominator.
$\endgroup$ 1 $\begingroup$Notice,
$\color{blue}{Method-1}: \text{L'Hospital's rule }$: Applying L'Hospital's rule for $\frac{0}{0}$ form as follows $$\lim_{x\to 13}\frac{5x-65}{169-x^2}$$ $$=\lim_{x\to 13}\frac{\frac{d}{dx}(5x-65)}{\frac{d}{dx}(169-x^2)}$$ $$=\lim_{x\to 13}\frac{5}{-2x}=-\frac{5}{2\cdot 13}=\color{blue}{-\frac{5}{26}}$$
$\color{red}{Method-2}: \text{Factorization}$: Applying $a^2-b^2=(a-b)(a+b)$ as follows $$\lim_{x\to 13}\frac{5x-65}{169-x^2}$$ $$=\lim_{x\to 13}\frac{5(x-13)}{13^2-x^2}$$ $$=\lim_{x\to 13}\frac{-5(13-x)}{(13-x)(13+x)}$$ $$=\lim_{x\to 13}\frac{-5}{13+x}=\frac{-5}{13+13}=\color{red}{-\frac{5}{26}}$$
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