Question

ww4: problem 25 (1 point) in this question, we will evaluate the limit $lim_{x
ightarrow1}\frac{x^{1838}-1}{x - 1}$. (a) is the function $\frac{x^{1838}-1}{x - 1}$ continuous at $x = 1$? what does that mean for computing the limit? the function is not continuous at x=1, so we need to simplify, or do something else, to decide whether the limit exists or not. (b) make the substitution $h=x - 1$. what happens to the limit? $lim_{x
ightarrow1}\frac{x^{1838}-1}{x - 1}=lim_{h
ightarrow0}1838$ (c) your limit in (b) looks a lot like a derivative. find a function $f(x)$ and a constant $a$ so that your limit above computes $f(a)$. then using derivative rules, evaluate the limit. $lim_{x
ightarrow1}\frac{x^{1838}-1}{x - 1}=f(a)=1838$ note: you can earn partial credit on this problem.

Explanation:

Step1: Analyze continuity

The function $\frac{x^{1838}-1}{x - 1}$ is not continuous at $x = 1$ since it is in $\frac{0}{0}$ form when $x=1$. We need alternative methods to find the limit.

Step2: Make substitution

Let $h=x - 1$, then $x=h + 1$. As $x
ightarrow1$, $h
ightarrow0$. So $\lim_{x
ightarrow1}\frac{x^{1838}-1}{x - 1}=\lim_{h
ightarrow0}\frac{(h + 1)^{1838}-1}{h}$. By the binomial expansion $(a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}$, when $a = 1$ and $b=h$, $(h + 1)^{1838}=1+1838h+\text{(higher - order terms of }h)$. Then $\frac{(h + 1)^{1838}-1}{h}=\frac{1+1838h+\text{(higher - order terms of }h)-1}{h}=\frac{1838h+\text{(higher - order terms of }h)}{h}=1838+\text{(terms with }h)$. As $h
ightarrow0$, the limit is 1838.

Step3: Relate to derivative

The limit definition of the derivative is $f^{\prime}(a)=\lim_{x
ightarrow a}\frac{f(x)-f(a)}{x - a}$. If we let $f(x)=x^{1838}$ and $a = 1$, then $f^{\prime}(x)=1838x^{1837}$ and $f^{\prime}(1)=1838\times1^{1837}=1838$. So $\lim_{x
ightarrow1}\frac{x^{1838}-1}{x - 1}=f^{\prime}(1)=1838$.

Answer:

1838