cj104
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We basically need to find lim x → −3 f(x), lim x → −3+ f(x) and lim x → −3 f(x).
Initially, I thought the answer would be 4 for all three but then again at this point there is a 'corner' which would mean the RHS and LHS don't match (& therefore the normal limit doesn't exist...?)
So if the side limits are not equal to 4 I don't understand how we can use the graph to find them. I thought of the left one being 0 and the right one being 1 but aren't they 'too far'?
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RDKGames
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(Original post by cj104)
We basically need to find lim x → −3 f(x), lim x → −3+ f(x) and lim x → −3 f(x).
Initially, I thought the answer would be 4 for all three but then again at this point there is a 'corner' which would mean the RHS and LHS don't match (& therefore the normal limit doesn't exist...?)
So if the side limits are not equal to 4 I don't understand how we can use the graph to find them. I thought of the left one being 0 and the right one being 1 but aren't they 'too far'?
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The limit exists, and you are correct with 4.

Due to the corner, what doesn't exist is the gradient (derivative!), since it's clearly different depending on which side you approach -3 from.
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cj104
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(Original post by RDKGames)
The limit exists, and you are correct with 4.

Due to the corner, what doesn't exist is the gradient (derivative!), since it's clearly different depending on which side you approach -3 from.
Looking at my notes, you're right about differentiability not existing when there is a corner. However, it also says we test this by finding that the side limits do not match. This isn't the case here if they're both equal to 4?
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RDKGames
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(Original post by cj104)
Looking at my notes, you're right about differentiability not existing when there is a corner. However, it also says we test this by finding that the side limits do not match. This isn't the case here if they're both equal to 4?
You are getting confused.

What you have in your question is finding \displaystyle \lim_{x \to -3} f(x) which is indeed 4.

What you seem to be confusing it with is the differentiability, which requires finding

f'(-3) = \displaystyle \lim_{h \to 0} \dfrac{f(-3+h) - f(-3)}{h}

... and it is *this* limit that does not exist because it is different depending on whether you approach -3 from left or right sides.
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cj104
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(Original post by RDKGames)
You are getting confused.

What you have in your question is finding \displaystyle \lim_{x \to -3} f(x) which is indeed 4.

What you seem to be confusing it with is the differentiability, which requires finding

f'(-3) = \displaystyle \lim_{h \to 0} \dfrac{f(-3+h) - f(-3)}{h}

... and it is *this* limit that does not exist because it is different depending on whether you approach -3 from left or right sides.
Thank you!
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