Question

Find the following limits or state that they do not exist. Assume \(a, b, c,\) and k are fixed real numbers. $$\lim _{h \rightarrow 0} \frac{\sqrt{16+h}-4}{h}$$

Short answer

Answer: The limit of the function as h approaches 0 is \(\frac{1}{8}\).

Step-by-step solution

Rationalize the numerator

Multiply and divide the expression by the conjugate of the numerator, which is \(\sqrt{16+h}+4\). $$ \lim _{h \rightarrow 0} \frac{\sqrt{16+h}-4}{h}\cdot\frac{\sqrt{16+h} + 4}{\sqrt{16+h} + 4} $$

Simplify the expression

After multiplying the numerators and denominators, simplify the expression. Notice that the numerator simplifies to the difference of squares: $$ \lim _{h \rightarrow 0} \frac{ (16+h)-(4)^2}{h(\sqrt{16+h}+4)} $$

Simplify further

Simplify the expression even more by canceling out the common terms and reducing the fraction: $$ \lim _{h \rightarrow 0} \frac{ h}{h(\sqrt{16+h}+4)} $$ $$ \lim _{h \rightarrow 0} \frac{ 1}{\sqrt{16+h}+4} $$

Find the limit

Plug in h=0 into the expression, to find the limit: $$ \lim _{h \rightarrow 0} \frac{ 1}{\sqrt{16+h}+4} = \frac{ 1}{\sqrt{16+0}+4} =\frac{1}{4+4} = \frac{1}{8} $$ The limit of the given function as h approaches 0 is \(\frac{1}{8}\).

Key concepts

Rationalizing the Numerator

In calculus, the technique of rationalizing the numerator is used when dealing with limits involving square roots. The goal is to eliminate the square root from the numerator to simplify the expression and make it easier to evaluate the limit. This is accomplished by multiplying both the numerator and the denominator by the conjugate of the numerator.

For instance, in the problem \( \lim _{h \rightarrow 0} \frac{\sqrt{16+h}-4}{h} \), the conjugate of \(\sqrt{16+h}-4\) is \(\sqrt{16+h}+4\). Multiplying by the conjugate results in a difference of squares, which simplifies to a polynomial that is often easier to manage. This process allows for the cancellation of the troublesome 'h' term in the denominator, leading to a solvable limit.

Difference of Squares

The difference of squares is a fundamental algebraic pattern given by \(a^2 - b^2 = (a+b)(a-b)\). In the context of limits, this property helps simplify expressions that contain square roots. After rationalizing the numerator, we exploit this pattern to transform the expression into one where the limit can be easily computed.

Let's look again at the expression from our limit problem. By multiplying the conjugate of the numerator, we turn \( (\sqrt{16+h})^2 - (4)^2 \) into a difference of squares, which simplifies the numerator to \(h\). This simplification is crucial; it cancels out the 'h' in the denominator, stripping the potential 0/0 indeterminate form and allowing us to directly evaluate the limit.

Limit of a Function

The limit of a function as \(h\) approaches a value describes what the function's output gets closer to as \(h\) gets arbitrarily close to that value, without necessarily reaching it. In many cases, a direct substitution can yield the limit, but not always. When direct substitution results in an indeterminate form like '0/0', we need another strategy, like rationalizing the numerator.

After applying previous algebraic techniques, the problem \( \lim _{h \rightarrow 0} \frac{\sqrt{16+h}-4}{h} \) simplifies to \( \lim _{h \rightarrow 0} \frac{1}{\sqrt{16+h}+4} \). Then, by substituting \(h = 0\) into the simplified expression, we find the limit to be \(\frac{1}{8}\). Understanding the limit concept is essential, as it's fundamental to the study of calculus, especially when assessing the behavior of functions near specific points.