The problem is given below:
f(x) = [Sec x – sqrt.2] / [x – π/4] if x ≠ π/4
= k if x = π/4
Is continuous at x = π/4
Find the value of “k”.
I do not get any idea to proceed.
Kindly enlighten me.
With warm regards,
Aranga
The problem is given below:
f(x) = [Sec x – sqrt.2] / [x – π/4] if x ≠ π/4
= k if x = π/4
Is continuous at x = π/4
Find the value of “k”.
I do not get any idea to proceed.
Kindly enlighten me.
With warm regards,
Aranga
Since you are asked to determine "k" so that the function is "continuous" start with the definition of continuous!
A function, f, is continuous at x= a if and only if $\displaystyle \lim_{x\to a}= f(a)$. So you need to determine $\displaystyle \lim_{x\to \frac{\pi}{4}}\frac{sec(x)- \sqrt{2}{x- \frac{\pi}{4}}$.
The difficulty, for me, is that how you do that depends upon what you know about "continuity", "limits", and "differentiability" of the trig functions. And I don't know that.
Simply setting x equal to $\displaystyle \frac{\pi}{4}$ gives $\displaystyle \frac{0}{0}$ so you might consider "L'hopital's rule". That, however requires that you know the derivative of sec(x) and if you know that sec(x) is differentiable at $\displaystyle \frac{\pi}{2}$ you could use the fact that this limit is the derivative sec(x) at $\displaystyle \frac{\pi}{4}$. If you can't use the differentiability of sec(x) you might rewrite the fraction as $\displaystyle \frac{\frac{1}{cos(x)}- \sqrt{2}}{x- \frac{\pi}{4}}= \frac{1}{cos(x)}\frac{1- \sqrt{2}cos(x)}{x- \frac{\pi}{4}}$.
I do not know L'Hospital's rule.
I am suppose to solve without that.
I came upto what you have mentioned. but after that I am not getting any idea to proceed.
In a similar model, I have seen {sinx - cos x}/[x- Pi/4]
I converted the numerator as sin [x-Pi/4].
But in the mentioned problem above, I am stuck.
help me further please.
Frankly I do not know how one proceeds without L'Hospital's rule. I say it is just busy work.
Do you know and use the mean value theorem?
Let $u = x - \dfrac{\pi}{4}$. As $x \to \dfrac{\pi}{4}, u \to 0$. So, we have:
$$\lim_{u \to 0} \dfrac{\sec\left( u + \dfrac{\pi}{4} \right) - \sqrt{2}}{u}$$
Multiply top and bottom by $\tan u$:
$$\lim_{u \to 0} \dfrac{\sin u}{u}\cdot \sec u \cdot \dfrac{\sec\left( u + \dfrac{\pi}{4} \right) - \sqrt{2} }{\tan u}$$
I don't recall where to go from there. I am guessing it has something to do with the sum of angles formula for secant, but I do not recall exactly.
I was wrong. Multiply top and bottom by $\sin u$:
$$\lim_{u \to 0} \dfrac{\sin u}{u} \cdot \dfrac{\sec \left(u + \dfrac{\pi}{4} \right) - \sqrt{2}}{\sin u}$$
$$\lim_{u \to 0} \dfrac{\sin u}{u} \cdot \dfrac{\dfrac{1}{\cos \left(u + \dfrac{\pi}{4} \right)} - \sqrt{2}}{\sin u}$$
$$\lim_{u \to 0} \dfrac{\sin u}{u} \cdot \dfrac{\dfrac{\sqrt{2}}{\cos u - \sin u} - \sqrt{2}}{\sin u}$$
$$\sqrt{2}\lim_{u \to 0} \dfrac{\sin u}{u} \cdot \dfrac{1-\cos u + \sin u}{\sin u(\cos u - \sin u)}$$
$$\sqrt{2}\lim_{u \to 0} \dfrac{\sin u}{u} \cdot \left(\dfrac{1-\cos u}{\sin u(\cos u - \sin u)} + \dfrac{ \sin u}{\sin u(\cos u - \sin u)}\right)$$
$$\sqrt{2}\lim_{u \to 0} \dfrac{\sin u}{u} \cdot \left(\dfrac{1-\cos u}{\sin u(\cos u - \sin u)}\cdot \dfrac{1+\cos u}{1 + \cos u} + \dfrac{1}{\cos u - \sin u}\right)$$
$$\sqrt{2}\lim_{u \to 0} \dfrac{\sin u}{u} \cdot \left(\dfrac{\sin u}{(\cos u - \sin u)(1+\cos u)} + \dfrac{1}{\cos u - \sin u}\right)$$
$$\sqrt{2}\left(\lim_{u \to 0} \dfrac{\sin u}{u} \right) \cdot \left( \lim_{u \to 0} \dfrac{\sin u}{(\cos u - \sin u)(1+\cos u)} + \dfrac{1}{\cos u - \sin u}\right)$$
$$\sqrt{2}(1) \cdot \left( \dfrac{0}{(1-0)(1+1)} + \dfrac{1}{1-0}\right) = \sqrt{2}$$