# Thread: composition of limits as x tends to infinity

1. ## composition of limits as x tends to infinity

Let f,g: R->R be functions. Suppose f(x)->k as x-> infinity and that g is continuous at k. Prove g(f(x))->g(k) as x -> inifnity

My attempt:

Assume not

Then |g(f(x)) - g(k)| > epsilon > 0 for all x > 0
so |g(k) - g(k)| > epsilon > 0 by continuity

which is a contradiction

Is this correct? Or am I missing something?

many thanks

2. From the continuity of $g$ at $k$ we get $\varepsilon > 0\; \Rightarrow \;\left( {\exists \delta > 0} \right)\left[ {\left| {y - k} \right| < \delta \, \Rightarrow \,\left| {g(y) - k} \right| < \varepsilon } \right]$.

From the given we also have $\left( {\exists N} \right)\left[ {x > N\, \Rightarrow \,\left| {f(x) - k} \right| < \delta } \right]$.

Put those two together to finish.