The limit I’m working with is:
$$\lim_{\lambda\to\infty} \left(-t\sqrt\lambda + \lambda e^{t/\sqrt\lambda} - \lambda \right)$$
Just applying the limit directly I guess we’d get: $-\infty$ because the dominating term is $-\lambda$. The second term gets multiplied by $e^{\frac{t}{\sqrt\lambda}}$ which goes to $0$ as $\lambda$ goes to $\infty$.
However, when we apply a Taylor expansion for $e^{\frac{t}{\sqrt\lambda}}$:
$$\lim_{\lambda\to\infty} \left(-t\sqrt\lambda + \lambda \left(e^0 + \frac{t}{\sqrt\lambda} +\frac{1}{2} \left(\frac{t}{\sqrt\lambda} \right)^2 +\frac{1}{3!} \left( \frac{t}{\sqrt\lambda} \right)^3 +\ldots \right) -\lambda \right).$$
Multiplying the Taylor expansion out by $\lambda$ and after applying $\lambda\to\infty$ we have all the terms after $\frac{1}{2} \left( \frac{t}{\sqrt\lambda} \right)^2$ go to zero. The terms directly before $\frac{1}{2} \left( \frac{t}{\sqrt \lambda} \right)^2$ are canceled out by terms $-t\sqrt\lambda$ and $- \lambda$.
Therefore, all we’re left with is $\frac{1}{2}t^2$.
I guess the second approach to this problem is the correct approach just because it uses more math (???). Can someone tell me why the first approach is incorrect? Or I guess in general, if someone can point out a way to approach unclear limits, that’d be incredibly helpful.
The error in the first approach is that $e^\frac{t}{\sqrt \lambda}$ doesn't go to $0$ as $\lambda$ goes to $\infty$, it goes to $1$. Thus both second and third term have order of growth of approximately $\lambda$.
Your second approach is correct, except you need to write a reminder for Taylor series (and fix typo in denominator): $e^\frac{t}{\sqrt \lambda} = 1 + \frac{t}{\sqrt \lambda} + \frac{t^2}{2 \lambda} + O\left(\frac{1}{\lambda^{3/2}}\right)$, which gives us after substitution $\frac{t^2}{2}$.