How to find the value of h and k in limit without L'Hôpital's rule?

Question

Given that $$ \lim_{x \rightarrow -2^-}\frac{x+2}{\sqrt{x+h}+k} = 6 $$ Find the value of h and k.

I have tried my best but the limit always $$ \lim_{x \rightarrow -2^-}\sqrt{x+h} - \sqrt{h-2} = 0 $$

Answer 1

You have that $$ \frac{x+2}{\sqrt{x+h}+k} = \frac{x+2}{x+h-k^2}(\sqrt{x+h}-k) $$ As $x\to -2$, this has a nonzero limit only if $h-k^2=2$. The limit is then $\sqrt{-2+h}-k$ so $(6+k)^2=-2+h$. The only solution is $k=-3$, $h=11$.

Answer 2

Let $x=y-2$ with $y \to 0$ then

$$\frac{x+2}{\sqrt{x+h}+k}=\frac{y}{\sqrt{y-2+h}+k}=\frac{y}{\sqrt{h-2}\sqrt{1+\frac{y}{h-2}}+k}\sim \frac{y}{\sqrt{h-2}\left(1+\frac{y}{2(h-2)}\right)+k}=\frac{2y(h-2)}{2(h-2)\sqrt{h-2}+y\sqrt{h-2}+2k(h-2)}$$

and this tends to $6$ if and only if

• $2(h-2)\sqrt{h-2}+2k(h-2)=0$
• $\frac{2(h-2)}{\sqrt{h-2}}=2\sqrt{h-2}=6\implies h=11$

and then

• $2(11-2)\sqrt{11-2}+2k(11-2)=54+18k=0\implies k=-3$

Note that for

$\sqrt{1+\frac{y}{h-2}}\sim 1+\frac{y}{2(h-2)}$

we have used the binomial expansion for $x\to 0 \implies (1+x)^a\sim 1+ax$.

Answer 3

Since the numerator vanishes at $-2$, in order that the limit is $6$ also the denominator must vanish at $-2$; now $$ \sqrt{-2+h}+k=0 $$ implies $k\le0$ and $-2+h=k^2$. Therefore $h=k^2+2$. Now it's just rationalizing: for $x\ne-2$, $$ \frac{x+2}{\sqrt{x+2+k^2}+k}= \frac{x+2}{x+2}(\sqrt{x+2+k^2}-k)=\sqrt{x+2+k^2}-k $$ so you get $\sqrt{k^2}-k=6$.

Can you finish? Remember that $k\le0$.