# Prove using formal definition if limx->∞ f(x)=-∞andc>0,then limx->∞ c f(x)=-∞ limx->a g(x)=∞andg(x)≤f(x)forx->a,then limx->a f(x)=∞

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You need to use delta-epsilon definition to prove that `lim_(x-gtoo)c*f(x) = -oo` , if c>0 and`lim_(x-gtoo) f(x) = -oo ` such that:

`|f(x) - (oo)| lt epsilon gt0 =gt |-oo|lt epsilon =gt ooltepsilon = oo`

Since `cgt0 =gt |cf(x) - (oo)| lt epsilon =gt |-oo|ltepsilon = oo`

**Hence, if you evaluate the limit `lim_(x-gtoo)c*f(x) =lim_(x-gtoo)c*lim_(x-gtoo)f(x) = c*(-oo) = -oo` .**

You need to prove that`lim(x-gta) f(x) = oo ` if `lim(x-gta) g(x) = oo` and `g(x)=ltf(x), ` hence you should use delta-epsilon definition such that:

If `lim(x-gta) g(x) = oo, then |g(x) - oo| lt epsilon` .

The problem prvides the information that `g(x)=ltf(x) =gt f(x) - g(x) gt= 0` Hence, using delta-epsilon definition yields:

`|f(x) - g(x) - oo| lt epsilon =gt |-oo| ltepsilon = infinity`

**Hence, using delta-epsilon definition you may verify that `lim(x-gta) f(x) = oo` if `lim(x-gta) g(x) = oo` and `g(x)=ltf(x)` .**