Calculating Limits Of Quotients As H Approaches 0 A Step-by-Step Guide

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This comprehensive guide delves into the fascinating realm of calculus, specifically focusing on the concept of limits. We will explore how to determine the limit of quotients as the variable h approaches 0. This is a fundamental concept in calculus, with applications ranging from finding derivatives to understanding the behavior of functions. We will tackle two specific examples, providing step-by-step solutions and explanations to solidify your understanding.

(a) $\frac{\sqrt{x+h}-\sqrt{x}}{h}$

Understanding the Limit Definition

In calculus, the limit of a function f(x) as x approaches a certain value c represents the value that f(x) gets arbitrarily close to as x gets arbitrarily close to c. In our case, we are interested in the limit of the given quotient as h approaches 0. This means we want to find out what value the expression $ rac{\sqrt{x+h}-\sqrt{x}}{h}$ approaches as h gets closer and closer to 0. This limit is particularly important because it forms the basis for the definition of the derivative.

Direct substitution of h = 0 into the expression results in an indeterminate form (0/0), which means we cannot directly evaluate the limit. To overcome this, we need to employ algebraic techniques to manipulate the expression and eliminate the indeterminate form. The key technique here is to rationalize the numerator.

Rationalizing the Numerator: A Step-by-Step Approach

Rationalizing the numerator involves multiplying both the numerator and the denominator of the fraction by the conjugate of the numerator. The conjugate of $\sqrt{x+h}-\sqrt{x}$ is $\sqrt{x+h}+\sqrt{x}$. This process eliminates the square roots in the numerator, making the expression easier to work with.

Let's apply this technique step-by-step:

  1. Multiply by the conjugate:

    x+h−xh∗x+h+xx+h+x\frac{\sqrt{x+h}-\sqrt{x}}{h} * \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}

  2. Expand the numerator: The numerator becomes a difference of squares: ((\sqrt{x+h})^2 - (\sqrt{x})^2), which simplifies to (x + h - x).

  3. Simplify the expression: The expression now looks like this:

    x+h−xh(x+h+x)\frac{x + h - x}{h(\sqrt{x+h}+\sqrt{x})}

    Simplifying the numerator, we get:

    hh(x+h+x)\frac{h}{h(\sqrt{x+h}+\sqrt{x})}

  4. Cancel out the common factor: We can cancel out the h in the numerator and denominator (since h is approaching 0 but is not equal to 0):

    1x+h+x\frac{1}{\sqrt{x+h}+\sqrt{x}}

Evaluating the Limit

Now that we have simplified the expression, we can directly substitute h = 0 to evaluate the limit:

lim⁡h→01x+h+x=1x+0+x=1x+x=12x\lim_{h \to 0} \frac{1}{\sqrt{x+h}+\sqrt{x}} = \frac{1}{\sqrt{x+0}+\sqrt{x}} = \frac{1}{\sqrt{x}+\sqrt{x}} = \frac{1}{2\sqrt{x}}

Therefore, the limit of the quotient $\frac{\sqrt{x+h}-\sqrt{x}}{h}$ as h approaches 0 is $\frac{1}{2\sqrt{x}}$. This result is significant because it represents the derivative of the function $\sqrt{x}$.

Connecting to Derivatives

It's crucial to understand the connection between limits and derivatives. The derivative of a function f(x) at a point x is defined as the limit:

f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}

In our case, f(x) = $\sqrt{x}$, and we have just calculated the limit, which gives us the derivative of $\sqrt{x}$, which is $\frac{1}{2\sqrt{x}}$. This highlights how the concept of limits is fundamental to defining derivatives in calculus.

(b) $\frac{\sin (x+h)-\sin x}{h}$

Applying Trigonometric Identities

This quotient involves trigonometric functions, specifically the sine function. Again, direct substitution of h = 0 results in the indeterminate form (0/0). To evaluate this limit, we need to employ trigonometric identities to manipulate the expression. The key identity we will use is the sine addition formula:

sin⁡(a+b)=sin⁡acos⁡b+cos⁡asin⁡b\sin(a + b) = \sin a \cos b + \cos a \sin b

Step-by-Step Solution

  1. Apply the sine addition formula: We can rewrite $\sin(x+h)$ using the sine addition formula:

    sin⁡(x+h)=sin⁡xcos⁡h+cos⁡xsin⁡h\sin(x+h) = \sin x \cos h + \cos x \sin h

  2. Substitute into the expression: Substituting this back into the original quotient, we get:

    sin⁡xcos⁡h+cos⁡xsin⁡h−sin⁡xh\frac{\sin x \cos h + \cos x \sin h - \sin x}{h}

  3. Rearrange the terms: Rearrange the terms in the numerator to group the $\sin x$ terms together:

    sin⁡x(cos⁡h−1)+cos⁡xsin⁡hh\frac{\sin x (\cos h - 1) + \cos x \sin h}{h}

  4. Separate the fraction: We can separate the fraction into two terms:

    sin⁡x(cos⁡h−1)h+cos⁡xsin⁡hh\frac{\sin x (\cos h - 1)}{h} + \frac{\cos x \sin h}{h}

    Which can be written as:

    sin⁡x∗(cos⁡h−1)h+cos⁡x∗sin⁡hh\sin x * \frac{(\cos h - 1)}{h} + \cos x * \frac{\sin h}{h}

Utilizing Known Limits

To evaluate the limit of this expression, we need to recall two crucial limits:

  • lim⁡h→0sin⁡hh=1\lim_{h \to 0} \frac{\sin h}{h} = 1

  • lim⁡h→0cos⁡h−1h=0\lim_{h \to 0} \frac{\cos h - 1}{h} = 0

These limits are fundamental in calculus and can be proven using geometric arguments or L'Hôpital's Rule.

Evaluating the Limit

Now we can evaluate the limit of the entire expression as h approaches 0:

lim⁡h→0[sin⁡x∗(cos⁡h−1)h+cos⁡x∗sin⁡hh]\lim_{h \to 0} [\sin x * \frac{(\cos h - 1)}{h} + \cos x * \frac{\sin h}{h}]

Using the known limits, we get:

sin⁡x∗lim⁡h→0(cos⁡h−1)h+cos⁡x∗lim⁡h→0sin⁡hh=sin⁡x∗0+cos⁡x∗1=cos⁡x\sin x * \lim_{h \to 0} \frac{(\cos h - 1)}{h} + \cos x * \lim_{h \to 0} \frac{\sin h}{h} = \sin x * 0 + \cos x * 1 = \cos x

Therefore, the limit of the quotient $\frac{\sin (x+h)-\sin x}{h}$ as h approaches 0 is $\cos x$.

Connection to the Derivative of Sine

Just like in the previous example, this limit has a direct connection to the derivative. The result we obtained, $\\cos x$, is precisely the derivative of the sine function. This reinforces the idea that limits are the foundation upon which derivatives are built.

The Derivative of sin(x)

As we've seen, the limit

lim⁡h→0sin⁡(x+h)−sin⁡(x)h\lim_{h \to 0} \frac{\sin(x+h) - \sin(x)}{h}

equals $\\cos(x)$. This is a crucial result in calculus, demonstrating that the derivative of the sine function is the cosine function. This relationship is fundamental to understanding the behavior of trigonometric functions in calculus and their applications in various fields, such as physics and engineering.

Conclusion

In this guide, we have meticulously explored the process of finding limits of quotients as h approaches 0. We tackled two distinct examples, each requiring different algebraic and trigonometric techniques. The first example involved rationalizing the numerator to eliminate the indeterminate form, while the second required the application of trigonometric identities and known limits. Both examples highlighted the fundamental connection between limits and derivatives, showcasing how limits form the bedrock of differential calculus. Mastering these techniques and understanding the underlying concepts is crucial for success in calculus and related fields. The ability to manipulate expressions, apply relevant identities, and utilize known limits is paramount to solving limit problems and grasping the essence of calculus.