Answer:
1
Step-by-step explanation:
Before we do that limit, let's look at:
[tex]\lim_{x \rightarrow \infty} \frac{-1}{x}[/tex]
[tex]-\lim_{x \rightarrow \infty} \frac{1}{x}=0[/tex]
You can look at a graph of [tex]y=\frac{1}{x}[/tex] if that convinces you. The graph approaches [tex]y=0[/tex] as [tex]x[/tex] gets larger and larger.
Let's get back to:
[tex]\lim_{x \rightarrow \infty} e^{\frac{-1}{x}}[/tex]
Let [tex]u=\frac{-1}{x}[/tex].
Then we have [tex]\lim_{u \rightarrow 0} e^{u}[/tex].
This function is actually continuous at [tex]u=0[/tex] so we can use direct substitution.
Conclusion:
[tex]\lim_{x \rightarrow \infty} e^{\frac{-1}{x}}=\lim_{u \rightarrow 0} e^{u}=e^0=1[/tex]
