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$$\lim_{x\to2} \frac{x^2-cx+d}{x^2-4} = 3$$ Find $c$ and $d$.

I tried replacing all the x's with 2, but ended up with 0 on the bottom. In order for the limit to exist, something from the top has to cancel out with $(x+2)$ or $(x-2)$. How do I find $c$ and $d$?

user
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Amanda Choi
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2 Answers2

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Your thought process is right. We do need to cancel out the $x-2$ in the bottom to not divide by zero. So assume the numerator is of the form $(x-2)(x+a)=x^2-cx+d$.

Then once we cancel everything out from the original limit, we have $$\lim_{x\to2} \frac{x+a}{x+2}=3$$

$\frac{2+a}4=3$ ---> $a=10$

Now go back and multiply everything out: $(x-2)(x+10) = x^2+8x-20$.

Then $c=-8, d=-20$.

Christopher Marley
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We have that

$$\lim_{x\to 2}\frac{x^2-cx+d}{x^2-4}=3$$

requires that for $x=2$

$$x^2-cx+d=0 \implies 4-2c+d=0 \implies d=-4+2c$$

then

$$x^2-cx+d=x^2-cx-4+2c=(x-2)(x+2)-c(x-2)=(x-2)(x+2-c)$$

then

$$\lim_{x\to 2}\frac{x^2-cx+d}{x^2-4}=\lim_{x\to 2}\frac{(x-2)(x+2-c)}{(x-2)(x+2)}$$

As an alternative once we recognize that the numerator must vanish at $x=2$, we can use l'Hopital to obtain

$$\lim_{x\to 2}\frac{x^2-cx+d}{x^2-4}=\lim_{x\to 2}\frac{2x-c}{2x}=3$$

from which we can find $c$ and then $d$.

user
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  • Just one critique, no need to introduce l'Hopital where unnecessary. Much too advanced for pre-cal. – Christopher Marley Nov 23 '18 at 22:58
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    @ChristopherMarley I agree and I don't like to use l'Hopital at all when it is not strictly necessary and indeed I'm not suggesting that as the first choice. But I think it could be a good alternative in that case at least to double check the result. – user Nov 23 '18 at 23:01