This is the progress I've made so far.

$$\lim_{h \to 0} \frac{\sqrt{1+e^{x+h}}-\sqrt{1+e^{x}}}{h}$$

$$= \lim_{h \to 0} \frac{\left(1+e^{x+h}\right)-\left(1+e^{x}\right)}{h\left(\sqrt{1+e^{x+h}}+\sqrt{1+e^{x}}\right)}$$

$$= \lim_{h \to 0} \frac{e^x\left(e^h-1\right)}{h\left(\sqrt{1+e^{x+h}}+\sqrt{1+e^{x}}\right)}$$

$$= \lim_{h \to 0} \frac{e^x\left(e^h-1\right)}{h\left(\sqrt{1+e^{x+h}}+\sqrt{1+e^{x}}\right)}$$

$$= \lim_{h \to 0} \frac{e^x}{\sqrt{1+e^{x+h}}+\sqrt{1+e^{x}}} \frac{e^h-1}{h}$$

I can see how as h tends to 0 the fraction on the left will tend to the desired result, but I'm not sure how to deal with the fraction on the right.

  • 1
    $\begingroup$ That's the derivative of $e^x$ at zero, which is $1$. $\endgroup$ – user296602 Apr 9 '16 at 16:51

In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequalities

$$1+x\le e^x\le \frac{1}{1-x} \tag 1$$

for $x<1$. Then, we have

$$1\le \frac{e^h-1}{h}\le \frac{1}{1-h}$$

whereupon applying the squeeze theorem, we obtain the coveted limit.

And in THIS ANSWER, I showed using the inequalities in $(1)$ that

$$\lim_{h\to 0}\frac{b^h-1}{h}=\log(b)$$

for $b>0$. Letting $b=e$ immediately gives

$$\lim_{h\to 0}\frac{e^h-1}{h}=1$$


hint use $\lim_{h\to 0}(1+h)^{1/h}=e$ so we can use this approximation and get $e^h \approx (1+h)$ and then substitute value of $h=0$ and get the answer as $1$!


$$\lim_{h \to 0} \frac{e^x}{\left(\sqrt{1+e^{x+h}}+\sqrt{1+e^{x}}\right)} \cdot \frac{\left(e^h-1\right)}{h}$$

Since $$e^h=\frac{h^0}{0!}+\frac{h}{1!}+\frac{h^2}{2!}+\frac{h^3}{3!}+......$$


$$e^h-1= \frac{h}{1!}+\frac{h^2}{2!}+\frac{h^3}{3!}+......$$

$$\frac{\left(e^h-1\right)}{h}= \frac{1}{1!}+\frac{h}{2!}+\frac{h^2}{3!}+......$$

Thus $$\lim_{h \to 0}\frac{\left(e^h-1\right)}{h}= 1$$

  • $\begingroup$ Using Taylor series assumes you know the derivative of $e^x$ which the OP isn't doing as he is using first principles. $\endgroup$ – Ian Miller Apr 9 '16 at 17:06
  • $\begingroup$ @IanMiller Yeah I suppose that he knows the derivative of $e^x$ and how to prove it using first principles. He needs $\sqrt{1+e^x}$ not $e^x$ $\endgroup$ – Angelo Mark Apr 9 '16 at 17:10

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