- #1

DarthRoni

- 31

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Hey there, I'm new to this forum. Today I thought I would brush up on my calculus.

I would just like to know if my method is correct. Is there an easier way to prove this ?

By the way, it's my first time using LaTeX, so bear with me.

I am trying to prove the following :

[tex]

\lim_{x\rightarrow 10} {x^2} = 100

[/tex]

So, I must find a δ in which the following holds

[tex]

\forall\epsilon>0\ \exists\ \delta>0\ such\ that\ 0<|x - 10| < \delta \implies |x^2 - 100|< \epsilon

[/tex]

I observe the following

[tex]|x-10| = |x + (-10)|[/tex] and by triangle inequality, [tex]|x|+|-10| > |x-10|[/tex] We will also note that [tex]|x| + |-10| = |x| + |10|\implies|x|+|10|>|x-10|[/tex]

Now we can find a δ in terms of ε. By reverse triangle inequality,

[tex]|x^2 - 100|\geq||x^2| - |100||\ and\ ||x^2| - |100||>|x^2| - |100|\ and \ |x^2| - |100| = |x|^2 - |10|^2[/tex][tex]\implies\epsilon > |x|^2 - |10|^2[/tex] Using this we can see that,[tex]\frac{\epsilon+|10|^2}{|x|}>|x|\ as\ well\ as \ \frac{\epsilon-|x|^2}{|10|^2}>|10| \ and\ since\ we\ know:\ |x|+|10| > |x-10|,[/tex]

[tex]\implies \frac{\epsilon+|10|^2}{|x|} + \frac{\epsilon-|x|^2}{|10|} > |x-10|[/tex]

So take [tex]\delta = \frac{\epsilon+|10|^2}{|x|} + \frac{\epsilon-|x|^2}{|10|}[/tex]

Is this correct? I would be really grateful if I got some feedback!

Can we also say the function is continuous [itex]\forall c\in\mathbb R[/itex] in the following way.

[tex]\lim_{x\rightarrow c} {x^2} = c^2[/tex] just take [tex]\delta = \frac{\epsilon+|c|^2}{|x|} + \frac{\epsilon-|x|^2}{|c|}[/tex]

I would just like to know if my method is correct. Is there an easier way to prove this ?

By the way, it's my first time using LaTeX, so bear with me.

I am trying to prove the following :

[tex]

\lim_{x\rightarrow 10} {x^2} = 100

[/tex]

So, I must find a δ in which the following holds

[tex]

\forall\epsilon>0\ \exists\ \delta>0\ such\ that\ 0<|x - 10| < \delta \implies |x^2 - 100|< \epsilon

[/tex]

I observe the following

[tex]|x-10| = |x + (-10)|[/tex] and by triangle inequality, [tex]|x|+|-10| > |x-10|[/tex] We will also note that [tex]|x| + |-10| = |x| + |10|\implies|x|+|10|>|x-10|[/tex]

Now we can find a δ in terms of ε. By reverse triangle inequality,

[tex]|x^2 - 100|\geq||x^2| - |100||\ and\ ||x^2| - |100||>|x^2| - |100|\ and \ |x^2| - |100| = |x|^2 - |10|^2[/tex][tex]\implies\epsilon > |x|^2 - |10|^2[/tex] Using this we can see that,[tex]\frac{\epsilon+|10|^2}{|x|}>|x|\ as\ well\ as \ \frac{\epsilon-|x|^2}{|10|^2}>|10| \ and\ since\ we\ know:\ |x|+|10| > |x-10|,[/tex]

[tex]\implies \frac{\epsilon+|10|^2}{|x|} + \frac{\epsilon-|x|^2}{|10|} > |x-10|[/tex]

So take [tex]\delta = \frac{\epsilon+|10|^2}{|x|} + \frac{\epsilon-|x|^2}{|10|}[/tex]

Is this correct? I would be really grateful if I got some feedback!

Can we also say the function is continuous [itex]\forall c\in\mathbb R[/itex] in the following way.

[tex]\lim_{x\rightarrow c} {x^2} = c^2[/tex] just take [tex]\delta = \frac{\epsilon+|c|^2}{|x|} + \frac{\epsilon-|x|^2}{|c|}[/tex]

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