Solve $\lim_{\theta\rightarrow\pi/4} \dfrac{\cos\theta-\frac{1}{\sqrt{2}}}{\theta-\frac{\pi}{4}}$ without L'Hopitals rule

Question

$$\lim_{\theta\rightarrow\pi/4} \dfrac{\cos\theta-\frac{1}{\sqrt{2}}}{\theta-\frac{\pi}{4}}$$

The suggested solution is: First substitute $h$ for the denominator and writhe $\theta$ in terms of $h$. Next, find the limit as $h$ approches zero. Note that $\lim \frac{\sin h}{h}=1$ and $\lim\frac{\cos h-1}{h}=0$.

Is there a way to solve this without L'hopitals rule? They use L'hopitals rule in the student solutions manual even though its not covered in any of the previous sections.

Answer 1

Another way without using the hint:

Recognize $\frac{1}{\sqrt 2}$ as $\cos \frac{\pi}{4}$ and use the identity $\cos(a)-\cos(b) = -2\sin\frac{a+b}{2} \ \sin\frac{a-b}{2}$ to get $$ \lim_{\theta \to\frac{\pi}{4}} -\frac{2\color{red}{\sin\left(\frac{\theta-\pi/4}{2}\right)} \sin\left(\frac{\pi/4+\theta}{2}\right)}{\color{red}{\frac{\theta-\pi/4}{2}} \times 2} $$ Now, use $\lim_{h\to 0} \frac{\sin h}{h}=1$ and you’ll be done.

Answer 2

Hint: Use $$\frac 1{\sqrt 2}=\frac{\sqrt 2}2=\cos\frac{\pi}4$$ Then use the formula for $\cos a-\cos b$

Answer 3

I'll try to get you started. Following the tip, let $h = \theta-\pi/4$. Then the expression becomes

$$\frac{\cos(\pi/4+h) -\frac{1}{\sqrt{2}}}{h}.$$

By sum of angles identity you get

$$\frac{ \cos(\pi/4)\cos h -\sin{\pi/4}\sin h -\frac{1}{\sqrt{2}}}{h}.$$

Now use the limits they gave you in the tip.

Answer 4

Using the given hints,

$$\lim_{\theta\rightarrow\pi/4} \dfrac{\cos\theta-\frac{1}{\sqrt{2}}}{\theta-\frac{\pi}{4}}=\lim_{h\to0}\frac{\cos(\frac\pi4+h)-\frac1{\sqrt2}}h \\=\lim_{h\to0}\frac{\frac1{\sqrt2}\cos h-\frac1{\sqrt2}\sin h-\frac1{\sqrt2}}h \\=\frac1{\sqrt2}\lim_{h\to0}\frac{\cos h-1}h-\frac1{\sqrt2}\lim_{h\to0}\frac{\sin h}h.$$

Answer 5

I think the tip is self-explanatory:

Substitute $h=\theta-\pi/4$, and you get $\theta=h+\pi/4$

With this substitution, the expression becomes:

$$\frac{\cos\theta-\frac{1}{\sqrt2}}{\theta-\frac{\pi}{4}}=\frac{\cos{(h+\frac{\pi}{4})}-\frac{1}{\sqrt2}}{h}$$

Use the sum of angles formula to get:

$$\lim_{h\rightarrow0}\frac{\cos(h)\cos(\frac{\pi}{4})-\sin(h)\sin(\frac{\pi}{4})-\frac{1}{\sqrt2}}{h}$$

With the identity $\cos(\frac{\pi}{4})=\sin(\frac{\pi}{4})=\frac{1}{\sqrt2}$, can you follow from here?