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$$\lim_{x\to0} \frac{f(e^{5x} - x^2) - f(1)}{x}$$

It is known that $f'(1) = -2$

Given this info, I'm left with many questions. I'm going to assume that I'll want to substitute for something. I'll let $g(x) = e^{5x} - x^2$. But how do I incorporate the fact that $f'(1) = -2$? Should $g(1) = f'(1)$?

Also, is the differentiation of g(x) as follows?: $$f'(e^{5x}-x^2) = 5e^{5x} - 2x $$ I figured this is wrong as it's inside the function notation.

user382540
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3 Answers3

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I would use L'hopital's rule: $$ \lim_{x\rightarrow 0}\frac{f(e^{5x}-x^2)-f(1)}{x}=\lim_{x\rightarrow 0}(5e^{5x}-2x)f'(e^{5x}-x^2)=5f'(1)=-10 $$ if $f'$ is continuous.

Alternatively, following your attempt, you could also define $g(x)=f(e^{5x}-x^2)$, and rewrite your expression as $$ \lim_{x\rightarrow 0}\frac{g(x)-g(0)}{x-0}=g'(0) $$ with $g'(x)=(5e^{5x}-2x)f'(e^{5x}-x^2)\Rightarrow g'(0)=5f'(1)=-10$

operatorerror
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$$\frac{f(e^{5x}-x^{2})-f(1)}{x}=\frac{f(1+e^{5x}-1-x^{2})-f(1)}{e^{5x}-1-x^{2}}\cdot\left(5\cdot\frac{e^{5x}-1}{5x}-x\right)$$ and this tends to $f'(1)\cdot 5=-10$ as $x\to 0$. The use of advanced tools like L'Hospital's Rule and Taylor series are mostly unnecessary for simple limit problems.

Paramanand Singh
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  • +1, but you still need L'Hôpital or Taylor to do $(e^{5x}-1)/5x \to1$, or at least to know that $(e^x)'=e^x $. – Martin Argerami Dec 07 '16 at 09:46
  • @MartinArgerami: The limit related to $e^{x}$ can be proved using any chosen definition of symbol $e^{x}$ without using L'Hospital's / Taylor. Thus for example we can easily show that $(e^{x} - 1)/x \to 1$ as $x \to 0$ if $e^{x}$ is defined as limit of sequence of $(1 + (x/n))^{n}$. See http://math.stackexchange.com/a/541330/72031 If we define $e^{x}$ via the exponential series then the limit is obvious. – Paramanand Singh Dec 07 '16 at 11:48
  • Could you further explain the second part? How did you get from the first equation to the second? – user382540 Dec 07 '16 at 16:12
  • @user382540: Multiply and divide the given fraction by $e^{5x}-1-x^{2}$ and then $(e^{5x}-1-x^{2})/x$ is expressed in a suitable form. – Paramanand Singh Dec 07 '16 at 16:45
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You could also use compose Taylor series to get $$f(e^{5x}-x^2)=f(1)+5 x f'(1)+x^2 \left(\frac{25 f''(1)}{2}+\frac{23 f'(1)}{2}\right)+O\left(x^3\right)$$ which makes $$\frac{f(e^{5x} - x^2) - f(1)}{x}=5 f'(1)+x \left(\frac{25 f''(1)}{2}+\frac{23 f'(1)}{2}\right)+O\left(x^2\right)$$

Claude Leibovici
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