Evaluate $\lim_{h \to 0} \dfrac{\sqrt{x + h} - \sqrt{x}}{h}$.

Evaluate limh → 0 dfrac√(x + h) - √(x)h. – Class 11 Mathematics

NCERT Exemplar · Q3

\lim_{h \to 0} \dfrac{\sqrt{x + h} - \sqrt{x}}{h}

Sikkim CbseShort· 3mImportance★★★★★est

56% · 98/175 Questions

✓ Free question

Concept understanding — Difference Quotient

The Difference Quotient: What It Is and Why It Matters

Imagine you're tracking the distance a car has travelled over time. At 2:00 PM, the odometer reads 40 km. At 2:30 PM, it reads 70 km. How fast was the car going on average during that half-hour?

You'd calculate: $\frac{70 - 40}{0.5 \text{ hours}} = 60$ km/h.

That fraction — change in distance divided by change in time — is the average rate of change. The difference quotient is just a formal, algebraic way of writing that same idea for any function.

The Intuition: Slope of a Secant Line

Take any function $f(x)$ . Pick two points on its graph: $(x, f(x))$ and $(x + h, f(x + h))$ , where $h$ is some horizontal step (positive or negative). The line that cuts through both points is called a secant line.

The slope of that secant line is:

\text{slope} = \frac{\text{rise}}{\text{run}} = \frac{f(x + h) - f(x)}{(x + h) - x} = \frac{f(x + h) - f(x)}{h}

That expression — $\frac{f(x + h) - f(x)}{h}$ — is the difference quotient.

Note

The name comes from "difference" (you subtract two function values) and "quotient" (you divide by $h$ ). It's literally a quotient of differences.

The Precise Statement

\frac{f(x + h) - f(x)}{h}, \quad h \neq 0

This gives the average rate of change of $f$ over the interval from $x$ to $x + h$ . Geometrically, it's the slope of the secant line through $(x, f(x))$ and $(x + h, f(x + h))$ .

Key restrictions:

$h$ cannot be zero (you can't divide by zero).
$x$ and $x + h$ must both be in the domain of $f$ .

A Concrete Example

Let $f(x) = x^2$ . Compute the difference quotient at $x = 3$ with $h = 0.1$ :

\frac{f(3 + 0.1) - f(3)}{0.1} = \frac{(3.1)^2 - 9}{0.1} = \frac{9.61 - 9}{0.1} = \frac{0.61}{0.1} = 6.1

This tells us: over the interval $[3, 3.1]$ , the function $x^2$ increases at an average rate of 6.1 units per unit change in $x$ .

If you shrink $h$ to 0.01, you'd get 6.01. As $h$ gets smaller, the average rate approaches 6 — which is exactly the instantaneous rate of change (the derivative) of $x^2$ at $x = 3$ .

Tip

The difference quotient is the bridge between average rates (which you can compute with simple algebra) and instantaneous rates (which require limits). When you take the limit as $h \to 0$ , you get the derivative.

Why You'll See It Everywhere

The difference quotient isn't just a classroom exercise. It's the foundation of calculus:

Derivatives: $\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$
Physics: average velocity → instantaneous velocity
Economics: average cost change → marginal cost
Any field that studies how things change

Every time you see a derivative, you're looking at the limit of a difference quotient. Master this one expression, and you've unlocked the core idea of differential calculus.

Watch out

A common mistake: forgetting that $h$ is the change in the input, not the output. The numerator $f(x+h) - f(x)$ is the change in the output. Keep them straight: $\frac{\Delta \text{output}}{\Delta \text{input}}$ .

The difference quotient is the direct precursor to the formal definition of a derivative in the NCERT Class 11 Mathematics chapter on Limits and Derivatives, and "difference quotient formula and examples" is a commonly searched topic for CBSE board and JEE Main preparation. Understanding it as the slope of a secant line is essential groundwork for the "limits and derivatives important questions" that build up to differentiation in Class 12.

To evaluate this limit, we first identify the indeterminate $\frac{0}{0}$ form, then rationalize the numerator to simplify the expression, leading to the result $\boxed{\frac{1}{2\sqrt{x}}}$ .

When evaluating limits, our first step is always to try direct substitution. If this yields a finite, defined value, that's our limit. However, often we encounter indeterminate forms like $\frac{0}{0}$ or $\frac{\infty}{\infty}$ . These forms do not tell us the limit directly; instead, they signal that algebraic manipulation is required before we can find the limit.

In this problem, we have an expression involving square roots in the numerator. When direct substitution leads to an indeterminate form like $\frac{0}{0}$ and square roots are present, a common and effective algebraic technique is rationalization. The goal of rationalization is to eliminate the square roots from either the numerator or the denominator by multiplying by the conjugate. This process often transforms the expression into a form where the problematic term (the one causing the $0$ in the denominator) can be cancelled out, allowing for direct substitution.

Let's evaluate the limit step by step.

Identify the Indeterminate Form: First, we attempt to substitute $h=0$ directly into the expression:

\frac{\sqrt{x + 0} - \sqrt{x}}{0} = \frac{\sqrt{x} - \sqrt{x}}{0} = \frac{0}{0}

This is an indeterminate form, meaning we cannot determine the limit by direct substitution alone. We need to algebraically manipulate the expression.

2. Choose the Algebraic Manipulation:

Since the numerator involves a difference of square roots, $\sqrt{x+h} - \sqrt{x}$ , rationalization is the appropriate technique. We will multiply both the numerator and the denominator by the conjugate of the numerator, which is $\sqrt{x+h} + \sqrt{x}$ . This uses the difference of squares identity: $(a-b)(a+b) = a^2 - b^2$ .

> [!FORMULA]
> The difference of squares identity is $a^2 - b^2 = (a-b)(a+b)$.

3. Perform Rationalization:

Multiply the numerator and denominator by the conjugate:

\lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h} \times \frac{\sqrt{x + h} + \sqrt{x}}{\sqrt{x + h} + \sqrt{x}}

Now, apply the difference of squares identity to the numerator, where $a = \sqrt{x+h}$ and $b = \sqrt{x}$:

(\sqrt{x + h} - \sqrt{x})(\sqrt{x + h} + \sqrt{x}) = (\sqrt{x + h})^2 - (\sqrt{x})^2 = (x + h) - x = h

So the expression becomes:

\lim_{h \to 0} \frac{h}{h(\sqrt{x + h} + \sqrt{x})}

Simplify the Expression: We can now cancel out the common factor $h$ from the numerator and the denominator. This is valid because as $h \to 0$ , $h$ is approaching $0$ but is not actually equal to $0$ .

\lim_{h \to 0} \frac{1}{\sqrt{x + h} + \sqrt{x}}

> [!WARNING]
> It is crucial to remember that we can only cancel out the $h$ term because we are taking a limit as $h \to 0$, not evaluating the expression at $h=0$. If $h$ were exactly $0$, the original expression would be undefined.

5. Evaluate the Limit:

Now that the indeterminate form has been removed, we can substitute $h=0$ into the simplified expression:

\frac{1}{\sqrt{x + 0} + \sqrt{x}} = \frac{1}{\sqrt{x} + \sqrt{x}} = \frac{1}{2\sqrt{x}}

> [!NOTE]
> This limit is a fundamental result in calculus. It represents the derivative of the function $f(x) = \sqrt{x}$ with respect to $x$, calculated from first principles. That is, $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$.

The value of the limit is $\frac{1}{2\sqrt{x}}$ .

✓Final answer

\boxed{\frac{1}{2\sqrt{x}}}

←Previous questionQ2 — Evaluate limx → (1)/(2) (4x² - 1)/(2x - 1).→Next questionQ4 — Evaluate limx → 0 dfrac(x + 2)(1)/(3) - 2(1)/(3)x.

Q.Evaluate lim⁡h→0x+h−xh\lim_{h \to 0} \dfrac{\sqrt{x + h} - \sqrt{x}}{h}limh→0​hx+h​−x​​.

🎓Unlock everything free for 14 days

Q.Evaluate $\lim_{h \to 0} \dfrac{\sqrt{x + h} - \sqrt{x}}{h}$ .

Unlock everything free for 14 days