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I’m attempting to solve a limit problem, and my current solution requires moving a limit inside the exponent. Symbolically, I’m attempting the following:

$$\lim_{x\rightarrow c}e^{\frac{f(x)}{g(x)}}=e^{\lim_{x\rightarrow c}\frac{f(x)}{g(x)}}$$

At this point the various functions are such that I can show that the limit of $\frac{f(x)}{g(x)}$ is $0$ by L’Hôspital’s Rule, and I can conclude that the limit of the overall expression is therefore $1$. But is this true? I know I can pull terms in and out of limits so long as those terms don’t rely on the variable with respect to which the limit is being evaluated, but can one move the limit inside the function and evaluate it this way? If yes, how do you prove that; if no, why?

DonielF
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1 Answers1

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Ok, I'll convert the above comment into an answer. First of all, note that $\dfrac{f(x)}{g(x)}$ has a finite limit as $x \to c$. Therefore, in a neighbourhood of $c$, say $U$ with $c \in U$, we know that $g(x) \neq 0$ since $\dfrac{f(x)}{g(x)}$ is certainly defined in such a neighbourhood.

Let $h : U \to \mathbb R$ be the following function : $$ h(x) = \begin{cases} e^\dfrac{f(x)}{g(x)} & x \neq c\\ e^{\lim_{x \to c} \frac{f(x)}{g(x)}} & x = c \end{cases} $$

Claim : $h$ is continuous at $c$.

Proof : Let $h_1 : U \to \mathbb R$ be given by $$h_1(x) = \begin{cases}\dfrac{f(x)}{g(x)} & x \neq c \\ \lim_{x \to c} \dfrac{f(x)}{g(x)} & x=c\end{cases}$$

Then we observe that :

  • $h_1$ is continuous at $c$ since $\lim_{x \to c} h_1(x) = h_1(c)$ exists.

  • $h(x) = e^{h_1(x)}$ for $x \in U$.

As the exponential is continuous at $h_1(c) = 0$ , the first bullet point and this fact combine to give that $h$ is continuous at $c$. $\blacksquare$

Finally, the definition of $h$ being continuous at $c$ is : $$ \lim_{x \to c} h(x) = h(c) $$

writing this out, $\lim_{x \to c} e^{\frac{f(x)}{g(x)}} = 1 = e^0 = e^{\lim_{x \to c} \dfrac{f(x)}{g(x)}}$.

Possible generalizations include the replacement of the exponential function $e^t$ by any function that is continuous at $c$.

Harley
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Sarvesh Ravichandran Iyer
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  • Just noticed this closed as a duplicate : I'd actually considered that particular post before I converted my earlier comment into an answer and spent some time over it, but I decided that the idea of "extending" $\frac{f(x)}{g(x)}$ in case it wasn't already defined at $c$, to be a continuous function at $c$ in order to apply the composition of continuity was too non-trivial to be included as part an answer to the duplicate post. @Harley thanks for the edit as well. – Sarvesh Ravichandran Iyer Oct 08 '21 at 01:24