Question

because of their connection with secant lines, tangents, and instantaneous rates, limits of the form $limlimits_{h\to 0} \frac{f(x+h)-f(x)}{h}$ occur frequently in calculus. evaluate this limit for the given value of x and function f. $f(x)=sqrt{x}, x=2$

Explanation:

Step1: Substitute \( f(x) \) and \( f(x + h) \)

Given \( f(x)=\sqrt{x} \), then \( f(x + h)=\sqrt{x + h} \). For \( x = 2 \), we have \( f(2)=\sqrt{2} \) and \( f(2 + h)=\sqrt{2 + h} \). Substitute into the limit formula:
\[
\lim_{h
ightarrow0}\frac{\sqrt{2 + h}-\sqrt{2}}{h}
\]

Step2: Rationalize the numerator

Multiply the numerator and denominator by the conjugate of the numerator, \( \sqrt{2 + h}+\sqrt{2} \):
\[
\lim_{h
ightarrow0}\frac{(\sqrt{2 + h}-\sqrt{2})(\sqrt{2 + h}+\sqrt{2})}{h(\sqrt{2 + h}+\sqrt{2})}
\]
Using the difference of squares \( (a - b)(a + b)=a^{2}-b^{2} \) in the numerator:
\[
\lim_{h
ightarrow0}\frac{(2 + h)-2}{h(\sqrt{2 + h}+\sqrt{2})}
\]

Step3: Simplify the numerator

Simplify the numerator: \( (2 + h)-2=h \). So the limit becomes:
\[
\lim_{h
ightarrow0}\frac{h}{h(\sqrt{2 + h}+\sqrt{2})}
\]

Step4: Cancel \( h \) and evaluate the limit

Cancel \( h \) (since \( h
eq0 \) when taking the limit as \( h
ightarrow0 \)):
\[
\lim_{h
ightarrow0}\frac{1}{\sqrt{2 + h}+\sqrt{2}}
\]
Now substitute \( h = 0 \):
\[
\frac{1}{\sqrt{2+0}+\sqrt{2}}=\frac{1}{2\sqrt{2}}
\]
Rationalize the denominator:
\[
\frac{1}{2\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{4}
\]

Answer:

\( \frac{\sqrt{2}}{4} \)