Let the function be defined by f(x)= cases x^2-1 |x-1| ,&x1

MATHEMATICS · Multi-part · AP · Calculus AB · Limits and Continuity · Limits and Continuity

Problem

Let the function

f

be defined by

f(x)=\begin{cases}\dfrac{x^2-1}{\ln|x-1|},&x<1,\\ 4k, &x=1,\\ \dfrac{\cos(\pi x)-1}{x-1},&x>1,\end{cases}

where

k

is a real constant. (a) Compute

\displaystyle\lim_{x\to 1^-}f(x)

and

\displaystyle\lim_{x\to 1^+}f(x)

. Show all intermediate steps. (b) Determine the value of

k

that makes

f

continuous at

x=1

. Justify your answer. (c) The function

g

is defined by

g(x)=\dfrac{x-1}{f(x)}

for all

x\ne 1

. Evaluate

\displaystyle\lim_{x\to 1}g(x)

, or explain why the limit does not exist, using your result from part (b).