Question

4.8 lhopitals rule

1. use lhopital to find the following limits.

a. $\lim_{h\to 0}\frac{(1 + h)^{-2}-1}{h}$
f.

Explanation:

Step1: Check the form of the limit

When \(h = 0\), the numerator \((1 + h)^{-2}-1=(1 + 0)^{-2}-1=1 - 1=0\) and the denominator \(h = 0\). So, it is in the \(\frac{0}{0}\) form, and L'Hopital's Rule can be applied.
L'Hopital's Rule states that if \(\lim_{x
ightarrow a}\frac{f(x)}{g(x)}\) is in the \(\frac{0}{0}\) or \(\frac{\pm\infty}{\pm\infty}\) form, then \(\lim_{x
ightarrow a}\frac{f(x)}{g(x)}=\lim_{x
ightarrow a}\frac{f^{\prime}(x)}{g^{\prime}(x)}\)

Let \(f(h)=(1 + h)^{-2}-1\) and \(g(h)=h\)

First, find the derivative of \(f(h)\):
Using the chain - rule, if \(y=(1 + h)^{-2}\), let \(u = 1+h\), then \(y = u^{-2}\). \(\frac{dy}{du}=-2u^{-3}\) and \(\frac{du}{dh}=1\). So \(f^{\prime}(h)=\frac{d}{dh}[(1 + h)^{-2}-1]=-2(1 + h)^{-3}\)
The derivative of \(g(h)\) is \(g^{\prime}(h) = 1\)

Step2: Apply L'Hopital's Rule

\(\lim_{h
ightarrow0}\frac{(1 + h)^{-2}-1}{h}=\lim_{h
ightarrow0}\frac{-2(1 + h)^{-3}}{1}\)

Step3: Evaluate the new limit

Substitute \(h = 0\) into \(\frac{-2(1 + h)^{-3}}{1}\)
When \(h = 0\), we have \(\frac{-2(1+0)^{-3}}{1}=-2\)

Answer:

\(-2\)