Animated Solution for Mathematics - Limits, Continuity and Differentiability: Suppose $f(x)$ is differentiable at $x=1$ and ${\lim }_{h\to 0}\frac{1}{h}f(1+h)=5$, then $f^{\prime }(1)$ equals

Imagine you are standing on the graph of a function $f(x)$ at the point $x=1$. You are looking at a limit, a mathematical telescope that lets you see what happens as you zoom in infinitely close to this point.

The problem gives us a tantalizing clue:

At first glance, this looks like a simple expression, but look closer at the denominator. As $h$ approaches zero, the denominator is shrinking into nothingness.

In the world of limits, a denominator approaching zero is a red flag. If the numerator were any constant value, the expression would explode to infinity.

But here, the limit is a nice, finite number: $5$. This tells us something profound about the numerator, $f(1+h)$.

For the ratio to remain finite, the numerator must also be shrinking to zero at the exact same rate. This is the birth of the indeterminate form $\frac{0}{0}$.

Now, we must invoke one of the most powerful tools in our calculus toolkit: the relationship between differentiability and continuity. The problem explicitly tells us that $f(x)$ is differentiable at $x=1$.

This is not just a label; it is a guarantee. A function that is differentiable at a point is, by necessity, continuous at that point.

This means that as we approach $x=1$, the value of the function $f(1+h)$ must settle down to the value of the function at the point itself, $f(1)$.

Since we established that the limit of the numerator ${\lim }_{h\to 0}f(1+h)$ must be zero to satisfy our limit condition, it follows logically that $f(1)=0$. We have just uncovered a hidden coordinate: the point $(1,0)$ lies on our curve.

With $f(1)=0$ in our pocket, we can now look at the formal definition of the derivative. The derivative $f^{\prime }(1)$ is defined as the limit of the slope of the secant line as the distance between points vanishes:

This is the heartbeat of calculus. Now, watch the magic happen.

We know $f(1)=0$. When we substitute this into our definition, the expression becomes:

The zero effectively disappears, leaving us with:

Look at that final expression. It is identical to the limit provided in the original problem statement!

We were told that ${\lim }_{h\to 0}\frac{f(1+h)}{h}=5$. By simply connecting the definition of the derivative to the given limit, we have arrived at the answer.

The derivative $f^{\prime }(1)=5$.

It is a beautiful moment of mathematical symmetry where the complex definition of a derivative collapses into the very information we were given at the start. You have successfully navigated the trap of the indeterminate form and used the fundamental properties of calculus to reveal the truth.