Chapter 4, Problem 7RCC

(a)

To determine

To state: The L’Hospitals rule and explain.

Explanation of Solution

The L’Hospital’s rule:

“Suppose f and g are differentiable functions with $g^{\prime }\left(x\right)=0$ on an open interval that contains a such that $\underset{x\to a}{\lim }f\left(x\right)=0$ and $\underset{x\to a}{\lim }g\left(x\right)=0$ or $\underset{x\to a}{\lim }f\left(x\right)=\infty$ and $\underset{x\to a}{\lim }g\left(x\right)=\infty$ then $\underset{x\to a}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{\lim }\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}$”.

It states that if a limit of a quotient of a function is in an indeterminate form, then the limit of the quotient of the function is same as the limit of quotient of the derivatives subject to condition that all limits exist.

(b)

To determine

To explain: The use of L’Hospital’s rule for the limit of product of the functions fg such that $f\to 0\text{ and }g\to \infty$ as $x\to a$.

Explanation of Solution

Quotient rule for limits: “Limit of the quotient is quotient of the limits”.

Calculation:

Rearrange the given function in quotient form and then use the Quotient rule.

$\begin{array}{c}\underset{x\to a}{\lim }\left[f\left(x\right)\cdot g\left(x\right)\right]=\underset{x\to a}{\lim }\frac{f\left(x\right)}{\frac{1}{g\left(x\right)}}\\ =\frac{\underset{x\to a}{\lim }f\left(x\right)}{\underset{x\to a}{\lim }\frac{1}{g\left(x\right)}}\\ =\frac{\underset{x\to a}{\lim }f\left(x\right)}{\frac{1}{\underset{x\to a}{\lim }g\left(x\right)}}\\ =\frac{0}{\frac{1}{\infty }}\end{array}$

On further simplification, $\underset{x\to a}{\lim }\left[f\left(x\right)\cdot g\left(x\right)\right]=\frac{0}{0}$, which is in an indeterminate form.

So on applying L’Hospitals rule, it is easy to evaluate the given limit.

(c)

To determine

To explain: The use of L’Hospital’s rule for the limit of difference of the functions f$-$g such that $f\to \infty \text{ and }g\to \infty$ as $x\to a$.

Explanation of Solution

Sum rule for limits: “Limit of the sum is sum of the limits”.

Calculation:

Rewrite the given function as follows.

$f\left(x\right)-g\left(x\right)=\left(\frac{1}{g\left(x\right)}-\frac{1}{f\left(x\right)}\right)\left(f\left(x\right)g\left(x\right)\right)$

Use the sum rule and find the limit as follows,

$\begin{array}{c}\underset{x\to a}{\lim }\left[f\left(x\right)-g\left(x\right)\right]=\underset{x\to a}{\lim }\left[\left(\frac{1}{g\left(x\right)}-\frac{1}{f\left(x\right)}\right)\left(f\left(x\right)g\left(x\right)\right)\right]\\ =\underset{x\to a}{\lim }\left(\frac{1}{g\left(x\right)}-\frac{1}{f\left(x\right)}\right)\underset{x\to a}{\lim }\left(f\left(x\right)g\left(x\right)\right)\\ =\left(\frac{1}{\underset{x\to a}{\lim }g\left(x\right)}-\frac{1}{\underset{x\to a}{\lim }f\left(x\right)}\right)\left(\underset{x\to a}{\lim }f\left(x\right)\underset{x\to a}{\lim }g\left(x\right)\right)\\ =\left(\frac{1}{\infty }-\frac{1}{\infty }\right)\left(\infty \cdot \infty \right)\end{array}$

On further simplification, $\underset{x\to a}{\lim }\left[f\left(x\right)-g\left(x\right)\right]=0\cdot \infty$, which is in an indeterminate form.

So on applying L’Hospitals rule, it is easy to evaluate the given limit.

(d)

To determine

To explain: The use of L’Hospital’s rule for ${\left[f\left(x\right)\right]}^{g\left(x\right)}$ such that $f\to 0\text{ and }g\to 0$ as $x\to a$.

Explanation of Solution

Calculation:

Find the limit as follows.

$\begin{array}{c}\ln \underset{x\to a}{\lim }\left[f{\left(x\right)}^{g\left(x\right)}\right]=\underset{x\to a}{\lim }\ln \left(f{\left(x\right)}^{g\left(x\right)}\right)\\ =\underset{x\to a}{\lim }g\left(x\right)\left[\ln \left(f\left(x\right)\right)\right]\\ =\underset{x\to a}{\lim }\frac{\ln \left(f\left(x\right)\right)}{\frac{1}{g\left(x\right)}}\\ =\frac{\ln \left(0\right)}{\frac{1}{0}}\end{array}$.

The value of the above limit is in an indeterminate form.

So on applying L’Hospitals rule, it is easy to evaluate the given limit.