Y 2+ 1 y 2+ 1 X X 0 2 3 -1 0 1 2 3 -1- -1+ Graph of f Graph...

y 2+ 1 y 2+ 1 X X 0 2 3 -1 0 1 2 3 -1- -1+ Graph of f Graph of g The graphs of the functions $f$ and $g$ are shown in the figures above. Which of the following statements is false? (A) $lim_{x o 1} f(x) = 0$ (B) $lim_{x o 2} g(x)$ does not exist. (C) $lim_{x o 1} (f(x)g(x + 1))$ does not exist. (D) $lim_{x o 1} (f(x + 1)g(x))$ exists.

Transcript

00:01 So graph of fx and graph of gx is given and we need to find the out of four statement we need to find the which one is the fourth statement.

00:10 So the first statement is limit x tends to 1 fx is equal to zero.

00:16 So first statement is limit x tends to 1 fx is equal to zero.

00:22 Now as we see in the graph with the left hand limit so this is the point.

00:28 So limit x tends to 1 is over here.

00:33 So the left -hand limit is equal to right -hand limit and left -hand limit and left -hand limit is equal to right -hand limit and is equals to 0.

00:44 So this statement is correct.

00:46 Limit at x tends to 1 is 0.

00:49 Now option b.

00:50 So option a is correct.

00:53 Now option b, option b says that limit x tends to 2, gx does not exist.

01:00 Here, as we see in the graph, this is the two, so this is the right hand side limit and this is the right hand side limit.

01:09 So here the left hand limit is equals to, left hand limit is equals to 1 and the right hand limit is equals to minus 1.

01:20 So this means that left hand limit is not equals to right hand limit, therefore limit does not exist.

01:28 Hence the option b is also correct.

01:31 Now let's see option d.

01:35 First let's see option d.

01:36 Option d is limit x tends to 1, f of x plus 1 into g x.

01:46 So by substitution method i can write f of 2 plus g of 1.

01:52 So this comes out to be so f of 2 is minus 1 and g of 1, g of 1 is 1...

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