Limit result is wrong compared with specific values asigned I have two expressions: $t_1$ and $t_2$. I want to calculate division $t_1 / t_2$ when the parameters $r_1 = r_2$. This condition $r_1 = r_2$ causes the denominators of $t_1$ and $t_2$ to be indeterminate. So to avoid this indeterminacy I suppose that $r_1$ is different from $r_2$. By making this assumption that $r_1$ is different from $r_2$ I get the division $t_1 / t_2$. And for this division $t_1 / t_2$ I apply the limit when $r_1$ tends to $r_2$. The limit when $r_1$ tends to $r_2$ is calculated due to the indeterminacy whose cause is $r_1 = r_2$. When calculating this limit, the expression $t_1 / t_2$ converges to $-1$. However, even though the expression converges to $-1$, when taking almost equal values ​of $r_1$ and $r_2$, the expression takes on a different value. This value when $r_1$ is almost equal to $r_2$ the expression is equal to $1.5$.
$$t_1=\frac{-\sqrt{r_1 r_2 \left(K_1^2 r_1 r_2+K_1 K_2 \left(r_1^2+r_2^2\right)+K_2^2 r_1
   r_2\right)}+K_1 r_1 r_2+K_2 r_1^2}{(r_1-r_2) (K_1 r_2+K_2 r_1)}$$
$$
t_2=\frac{\sqrt{r_1 r_2 \left(K_1^2 r_1
   r_2+K_1 K_2
   \left(r_1^2+r_2^2\right)+K_2^2 r_1
   r_2\right)}-K_1 r_2^2-K_2 r_1
   r_2}{(r_1-r_2) (K_1 r_2+K_2
   r_1)}
$$ 
$$
\lim_{r_1\to r_2}\frac{t_1}{t_2}=-1
$$
But when assigning values, for example $k_1=20, k_2=30, r_1=0.1, r_2= 0.100001$, the value $t_1/t_2$ is equal to $1.5$
This for me is counterintuitive because the limit is $-1$ but when using values the expression does not get closer to $-1$ and is rather $1.5$ 
What could I improve? I want find the real limit value, that in this case is $1.5$
 A: You have:
$$\frac{t_1}{t_2}=-\frac{\sqrt{r_1 r_2 \left(K_1^2 r_1 r_2+K_1 K_2 \left(r_1^2+r_2^2\right)+K_2^2 r_1
   r_2\right)}-K_1 r_1 r_2-K_2 r_1^2}{\sqrt{r_1 r_2 \left(K_1^2 r_1 r_2+K_1 K_2 \left(r_1^2+r_2^2\right)+K_2^2 r_1
   r_2\right)}-K_2 r_1 r_2-K_1 r_2^2}$$
so with $r_1=r$ and $r_2=r+h$ your question is the limit when $h$ goes to $0$ of:
$$f(h)=-\frac{\sqrt{r(r+h)\left((k_1^2+k_2^2) r(r+h)+k_1 k_2(r^2+(r+h)^2) \right)}-k_1r(r+h)-k_2r^2}{\sqrt{r(r+h)\left((k_1^2+k_2^2) r(r+h)+k_1 k_2(r^2+(r+h)^2) \right)}-k_2r(r+h)-k_1(r+h)^2}$$
and:
$$\sqrt{r(r+h)\left((k_1^2+k_2^2) r(r+h)+k_1 k_2(r^2+(r+h)^2) \right)}-k_1r(r+h)-k_2r^2=\sqrt{(k_1+k_2)^2 r^4+2 h r^3 (k_1+k_2)^2+o(h)}-(k_1+k_2)r^2-k_1 r h=(k_1+k_2) r^2\left(\sqrt{1+\frac{2h}{r}+o(h/r)}-1 -\frac{k_1}{k_1+k_2}\frac{h}{r}+o(h/r)\right)=(k_1+k_2) r^2\left( 1+\frac{1}{2} \frac{2h}{r}-1- \frac{k_1}{k_1+k_2}\frac{h}{r}+o(h/r)\right)=\frac{k_2}{k_1+k_2} hr (1+o(1))$$
similarly:
$$\sqrt{r(r+h)\left((k_1^2+k_2^2) r(r+h)+k_1 k_2(r^2+(r+h)^2) \right)}-k_2r(r+h)-k_2(r+h)^2=\sqrt{(k_1+k_2)^2 r^4+2 h r^3 (k_1+k_2)^2+o(h)}-(k_1+k_2)r^2-k_2 r h-2 k_1 rh +r^2 o(h)=(k_1+k_2) r^2\left(\sqrt{1+\frac{2h}{r}+o(h/r)}-1 -\frac{k_2+2k_1}{k_1+k_2}\frac{h}{r}+o(h/r)\right)=(k_1+k_2) r^2\left( 1+\frac{1}{2} \frac{2h}{r}-1- \frac{k_2+2k_1}{k_1+k_2}\frac{h}{r}+o(h/r)\right)=\frac{-k_1}{k_1+k_2} hr (1+o(1))$$therefore:
$$f(h)=\frac{k_2/(k_1+k_2)}{k_1/(k_1+k_2)}\frac{1+o(1)}{1+o(1)}$$
and:
$$\lim_{h \to 0} f(h)=\frac{k_2}{k_1}$$
In your case $$\frac{k_2}{k_1}=\frac{30}{20}=1.5$$ which is the numerical result you obtained.

Remark:
In case you are not familiar with $o$ it is just a cleaner (and more powerful) way to use L'Hopital's rule.
A: When $r_1=r_2=r$ the numerators go to $0$ as well.  The square root becomes $(K_1+K_2)r^2$ and cancels with the other terms.  I would write the expressions using $r=r_1, d=r_2-r_1$ and take the limit as $d \to 0$.  The two denominators are the same, so divide out analytically.  Then letting $n_2$ be the numerator of $t_2$ and keeping only terms to order $d$ we have 
$$n_2=\sqrt{r_1 r_2 \left(K_1^2 r_1
   r_2+K_1 K_2
   \left(r_1^2+r_2^2\right)+K_2^2 r_1
   r_2\right)}-K_1 r_2^2-K_2 r_1
   r_2\\
=\sqrt{r^2(1+d)((K_1^2+K_2^2)r^2(1+d)+K_1K_2r^2(1+(1+d)^2))}-K_1r^2-K_2r^2(1+d)\\
=r^2\left(\sqrt{K_1^2+2K_1K_2+K_2^2+d(2K_1^2+4K_1K_2+2K_2^2)}-K_1-K_2-dK_2\right)\\
=r^2\left((K_1+K_2)(1+d)-K_1-K_2-dK_2\right)\\
=-r^2dK_2$$
I believe a similar simplification will happen for the first numerator and the $1.5$ is the ratio $\frac {K_2}{K_1}.
