Jtdylktuy

This is a question from Calculus 7e by Stewart:

It's easy to prove by calculating the limit of $e^{ln(f(x))cdot g(x)}$ if $f(x)$ is restricted to a positive function. However, seems that if we remove this restriction the conclusion still holds. But then we cannot use $ln(f(x))$ since $f(x)leq 0$ is possible. So how to prove the general form of this limit?

In calculus context for function of real variables, the function

$$[f(x)]^{g(x)}$$

is restricted and defined only for $f(x)>0$.

Refer also to the related

• Finding domain of $ f(x) ^ {g(x)} $?

In the given case, if we remove the restriction on f to be $geq0$, then $f(x)^{g(x)}$ cannot be defined(because the domain of this function is f(x)>0).

Reason for this is that g(x) may take fractional value that may not be defined for $f(x)^{g(x)}$(for example, square root of negative numbers are undefined)

Hope it helps:)

Let $varepsilon>0$. Then $minleft{varepsilon,frac12right}>0$ and therefore there is a $delta_1>0$ such that $$0<lvert x-arvert<delta_1impliesbigllvert f(x)bigrrvert<minleft{varepsilon,frac12right}.$$And there is a $delta_2>0$ such that$$0<lvert x-arvert<delta_2implies g(x)>1.$$Therefore, if $delta=min{delta_1,delta_2}$,$$0<lvert x-arvert<deltaimpliesleftlvert f(x)^{g(x)}rightrvert=bigllvert f(x)bigrrvert^{g(x)}<minleft{varepsilon,frac12right}leqslantvarepsilon.$$