Question

Let $y=f(x)$ be a function and let $P=(a, f(a))$ and $Q(a+h, f(a+h))$ be two points close to each other on the given graph of this function.
I. $\displaystyle\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}=\displaystyle\lim _{Q \rightarrow P} \frac{Q R}{P R}$
II. $f^{\prime}(a)=\tan \Psi$

Which of the following is/are true?

• A. Both I and II are true
• B. Only I is true
• C. Only II is true
• D. Both I and II are false

Answer 1

Answer: (A)

We know that, $f^{\prime}(a)=\displaystyle\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$
From the $\triangle P Q R$, it is clear that the ratio whose limit we are taking is precisely equal to $\tan (Q P R)$ which is the slope of the chord $P Q$. In the limiting process, as $h$ tends to 0 , the point $Q$ tends to $P$ and we have
$\displaystyle\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}=\displaystyle\lim _{Q \rightarrow P} \frac{Q R}{P R}$
This is equivalent to the fact that the chord $P Q$ tends to the tangent at $P$ of the curve $y=f(x)$. Thus, the limit turns out to be equal to the slope of the tangent. Hence,
$f^{\prime}(a)=\tan \psi .$