$T_2^4-T_1^4\approx4T_1^3(T_2-T_1)$ if $(T_2-T_1)/T_1$ small

Question

I read, in a text of elementary physics, that $$T_2^4-T_1^4\approx4T_1^3(T_2-T_1)$$ if $\Delta T:=T_2-T_1$ "is small with respect to $T_1$". In a rigourous language, I suppose that it means that $$\lim_{|\Delta T/T_1|\to 0} \frac{T_2^4-T_1^4}{4T_1^3(T_2-T_1)}=1$$but, please, correct me if I am wrong. Anyhow, I have not been able to handle the expression $\frac{T_2^4-T_1^4}{4T_1^3(T_2-T_1)}={(T_2^2+T_1^2)(T_2^2-T_1^2)}{4T_1^3(T_2-T_1)}$ in order to evidence the dependence from $\frac{T_2-T_1}{T_1}$. How can the validity of such an approximation proved? Thank you very much for any answer!

Answer 1

Notice $\frac{T_2 - T_1}{T_1} \to 0$ implies $T_2/T_1 \to 1$.Therefore,

$$\frac{T_2^4 - T_1^4}{4T_1^3(T_2 - T_1)} = \frac{(T_2^2 + T_1^2)(T_2 + T_1)}{4T_1^3}= \frac{1}{4}\left(1 + \left(\frac{T_2}{T_1}\right)^2\right)\left(1 + \frac{T_2}{T_1}\right) \to 1.$$ as $T_2/T_1 \to 1$.