Limited!

Calculus Level 4

Read the statements carefully.

$[1]$ . If $f(x)<g(x)$ for all $x$ , then $\displaystyle \lim_{x\to c} f(x) < \displaystyle \lim_{x\to c} g(x)$ provided that these limits exist.

$[2]$ . If both $\displaystyle \lim_{x\to c} f(x)$ and $\displaystyle \lim_{x\to c} g(x)$ do not exist, then it is impossible for $\displaystyle \lim_{x\to c} (f(x)+g(x))$ to exist.

$[3]$ . If $\displaystyle \lim_{x\to c} f(x)=l$ and $\displaystyle \lim_{x\to c} f(x)=m$ , then $l$ is always equal to $m$ .

Which of these are true?

This problem is from the set "MCQ Is Not As Easy As 1-2-3". You can see the rest of the problems here .

None of them are true. Only

[3]

All of them are true.

[1]

and

[3]

1 solution

Patrick Corn
Oct 29, 2014

[1] is false: $f(x) = 0$ , $g(x) = x^2$ for $x \ne 0$ and $1$ if $x = 0$ , $c = 0$ .

[2] is false: $f(x) = \sin(1/x)$ , $g(x) = 1-\sin(1/x)$ , $c=0$ .

[3] is trivial.

@Mursalin Habib What would be the case regarding [3] when the limit is an infinite number ?

Nishant Sharma - 6 years, 7 months ago

for third , firstly limit should exists

U Z - 6 years, 4 months ago

How can you say that statement (1) is false. I think it should be true. Your solution does not justify that it is false. Would you explain it a little more. From the functions you have assumed, $\displaystyle lim_{x \rightarrow 0}f(x) < \ lim_{x \rightarrow 0} g(x)$ . so how is it possible ?

Sandeep Bhardwaj - 6 years, 7 months ago

$\displaystyle \lim_{x \to 0} f(x) = \lim_{x \to 0} g(x) = 0$ . The limit of f is equal to the limit of g, not less than.

Ivan Koswara - 6 years, 4 months ago

Yes the limits are equal but the functions are also equal at this point! Those two functions don't satisfy the question!

I'm pretty sure 1 is true. If it were not, then the comparison test for infinite series wouldn't make any sense.

A Former Brilliant Member - 6 years, 4 months ago

is number 1 should be true???

Figel Ilham - 6 years, 6 months ago

I think the statement 1 should be true.

Prakhar Gupta - 6 years, 4 months ago

I think so, too.

Nick Lee - 6 years, 4 months ago

Limited!

This problem is from the set "MCQ Is Not As Easy As 1-2-3". You can see the rest of the problems here .

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