calculus - One sided limit with $e^x$ and $sin(x)$

Assuming that $e^{x} \to e^{a}$ and $\sin(x) \to \sin(a)$

evaluate the limit as

$$\lim_{x \to 0^{+}} \frac{e^{x^2+2x-1}}{\sin(x)}$$

To me, based on the first sentence. I could just plug in $a$ and get my limit which does not exist at the point $x=0$ However the back of the book says infinity so I am at a loss as to how to prove this.

This is a real analysis class but this is isn't one of the problems requiring us to prove with definitions but it would be helpful to learn the algebra and the theory. I don't understand as to how I can apply the definition of a function converging to infinity to prove this. Plus I don't see how that could help? How am I supposed to know it goes to infinity

Answer

When $x$ is close enough to $0$, then $-1.1 so that

$$\exp(-1.1)<\exp(x^2+2x-1)$$

and hence the function in your limit is bounded from below by $\frac{\exp(-1.1)}{\sin x}$ which blows up at $x=0$ since $\sin0=0$.