Composition of functions

**ManicGuy** · January 26th 2009, 12:46 AM

If we assume:

$ \begin{gathered} f\left( x \right) \to y_0 {\text{ as }}x \to x_0 \hfill \\ g\left( y \right) \to l{\text{ as }}y \to y_0 \hfill \\ g\left( {y_0 } \right) = l \hfill \\ \end{gathered} $ ,

how can I show that $ \mathop {\lim }\limits_{x \to x_0 } g\left( {f\left( x \right)} \right) = l $ ?

**ManicGuy** · January 26th 2009, 11:23 AM

Any ideas? I'm really stuck here.

**Plato** · January 26th 2009, 01:13 PM

If $\varepsilon > 0$ from the given $\left( {\exists \delta > 0} \right)\left[ {0 < \left| {y - y_0 } \right| < \delta \, \Rightarrow \,\left| {g(y) - l} \right| < \varepsilon } \right]$ .
Again $\,\delta > 0 \Rightarrow \,\left( {\exists \delta ' > 0} \right)\left[ {0 < \left| {x - x_0 } \right| < \delta '\, \Rightarrow \,\left| {f(x) - y_0 } \right| < \delta } \right]$ .
We are given that $g\left( {y_0 } \right) = l$ . Define $\delta '' = \min \left\{ {\delta ,\delta '} \right\}$ .

$0 < \left| {y - x_0 } \right| < \delta ''\, \Rightarrow \,\left| {f(y) - y_0 } \right| < \delta \, \Rightarrow \,\left| {g\left( {f(y)} \right) - l} \right| <\varepsilon$

Therefore: $g \circ f(y) \to l\;,\,y \to x_0$

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