Let's do some calculus! (31)

Calculus Level 4

$f(x) = \begin{cases} x+1 & x>0 \\ 2-x & x \le 0 \end{cases} \quad \quad g(x) = \begin{cases} x+3 & x<1 \\ x^2 - 2x - 2 & 1 \le x < 2 \\ x-5 & x \ge 2 \end{cases}$

Let $f(x)$ and $g(x)$ be functions satisfying the conditions above. Find $\displaystyle \left| \lim_{x \to 0} \big(g \circ f(x)\big) \right|$ .

Notations:

$g \circ f(x)$ denotes the function $g~\text{of}~ f(x)$ or $g\big(f(x)\big)$ .
$| \cdot |$ denotes the absolute value function .

For more problems on calculus, click here .

The answer is 3.

2 solutions

Chew-Seong Cheong
Sep 25, 2016

For $x > 0$ , $f(x) = x + 1 > 1$ then we have:

$\begin{aligned} g(f(x)) & = (f(x))^2 - 2f(x) - 2 & 1 \le x < 2 \\ & = (x+1)^2 - 2(x+1) - 2 \\ & = x^2+2x+1-2x-2-2 \\ & = x^2 - 3 \end{aligned}$

$\begin{aligned} \implies \lim_{x \to 0^+} g(f(x)) & = - 3 \end{aligned}$

For $x \le 0$ , $f(x) = 2-x \ge 2$ then we have:

$\begin{aligned} g(f(x)) & = f(x) - 5 & x \ge 2 \\ & = (2-x) - 5 \\ & = -x-3 \end{aligned}$

$\begin{aligned} \implies \lim_{x \to 0^-} g(f(x)) & = - 3 \end{aligned}$

Therefore,

$\begin{aligned} \lim_{x \to 0} g(f(x)) & = \lim_{x \to 0^+} g(f(x)) = \lim_{x \to 0^-} g(f(x)) = - 3 \\ \implies \left| \lim_{x \to 0} g(f(x))\right| & = |-3| = \boxed{3} \end{aligned}$

Viki Zeta
Sep 25, 2016

When $x=0$

$f(0) = 2-0=2$

$g(2)=2-5=-3$

Therefore the answer is |-3|=3

You should consider both Left hand limit and Right hand limits for your answer. What if both were different? Then there surely wouldn't have been limit.

Tapas Mazumdar - 4 years, 8 months ago

Let's do some calculus! (31)

For more problems on calculus, click here .

The answer is 3.

2 solutions

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