Proove that the only real constant such that the limit is $c=2$

Question

Prove that the only real constant such that the limit $$\lim_{h\rightarrow 0} {\tan(\frac{\pi}4+h) -1 -ch \over h}=0.$$ is $c=2$.

This is what I have tried so far L'Hôpital rule $$x = {\sin^2\frac{\pi}4+ \sin^2h -1 -ch\over 1}.$$ $$x = {1 -ch\over 1}.$$

I cannot figure out where I am going to find that constant $c$ is equal to $2$.

Answer 1

The application of l'Hopital Rule is a nice method but note that with respect to $h$

• the derivative of $-1$ is $0$
• the derivative of $-ch$ is $-c$
• and the derivative of $\tan\left(\frac{\pi}{4}+h\right)$ is $\frac{1}{\cos^2 \left(\frac{\pi}4 +h\right)}$

As an alternative note that by definition of derivative

$$\lim_{h\rightarrow 0}\frac{\tan( \frac{\pi}4\ +\ h)-1}{h}=\left(\tan x\right)'_{x=\frac{\pi}4}=\frac{1}{\cos^2 \frac{\pi}4 }=2$$

then

$$\lim_{h\rightarrow 0} \left(\frac{\tan\left( \frac{\pi}4\ +\ h\right) \ -1\ -ch}{h}\right) =\lim_{h\rightarrow 0} \left(\frac{\tan\left( \frac{\pi}4\ +\ h\right) \ -1}{h}-c\right) =0 \iff c=2$$

Answer 2

The faster way would be to use gimusi's approach but I will intentionally use L'hopital's rule

$$\lim\limits _{h\rightarrow 0}\left(\frac{\tan( \frac{\pi}4\ +\ h) \ -1\ -ch}{h}\right) =0$$

$$\lim\limits _{h\rightarrow 0}\left(\frac{\sec^2(\frac{\pi}4+h) -c}{1}\right) =0$$

Hence $$c = \lim_{h \to 0} \sec^2\left(\frac{\pi}{4}+h\right)=2$$