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powerless
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I started learning limits and I have some difficulties qith this question;
Find the limits or say why they do not exist.
a)[tex]lim_{x\rightarrow2}\frac{\left|x^2 + x -6\right|}{x-2}[/tex]
b) [tex]lim_{x\rightarrow0}\frac{\sqrt{4-x}-2}{\sqrt{x+1}-1}[/tex]
3. The Attempt at a Solution
a)we have an absolute value on the numerator but we can factorise it:
[tex]lim_{x\rightarrow2}\frac{\left|x^2 + x -6\right|}{x-2}[/tex]
[tex]lim_{x\rightarrow2}\frac{(x-2)(x+3)}{x-2}[/tex]
Now we have two cases:
(x approaching 2 from the positive side)
[tex]lim_{x\rightarrow2+}\frac{(x-2)(x+3)}{x-2}[/tex]
[tex]lim_{x\rightarrow2+}\frac{(2-2)(2+3)}{2-2}[/tex] = undefined (?)
(x approaching 2 from the negative side)
[tex]lim_{x\rightarrow2-}\frac{(-x+2)(x+3)}{x-2}[/tex]
[tex]lim_{x\rightarrow2-}\frac{(-2+2)(2+3)}{2-2}[/tex] = undefined
Why do I always get devision by zero and undefined, did I do somethin wrong?
b) [tex]lim_{x\rightarrow0}\frac{\sqrt{4-x}-2}{\sqrt{x+1}-1}[/tex]
Because we have a sqrt on the denominator we mulitply top & bottom by the conjugate and we rationalize the denominator,
[tex]lim_{x\rightarrow0}\frac{(\sqrt{4-x}-2)(\sqrt{x+1}+1)}{(\sqrt{x+1}-1)\sqrt{x+1}+1)}[/tex]
The denominator expands and simplifies to:
[tex]lim_{x\rightarrow0}\frac{(\sqrt{4-x}-2)(\sqrt{x+1}+1)}{x+1-1}[/tex]
What can i do next? should I expand the numerator too? ( it would become
[tex]\sqrt{4-x.x+1}+\sqrt{4-x}-2\sqrt{x+1}-2[/tex])
But I don't know how to do this one.
Find the limits or say why they do not exist.
a)[tex]lim_{x\rightarrow2}\frac{\left|x^2 + x -6\right|}{x-2}[/tex]
b) [tex]lim_{x\rightarrow0}\frac{\sqrt{4-x}-2}{\sqrt{x+1}-1}[/tex]
3. The Attempt at a Solution
a)we have an absolute value on the numerator but we can factorise it:
[tex]lim_{x\rightarrow2}\frac{\left|x^2 + x -6\right|}{x-2}[/tex]
[tex]lim_{x\rightarrow2}\frac{(x-2)(x+3)}{x-2}[/tex]
Now we have two cases:
(x approaching 2 from the positive side)
[tex]lim_{x\rightarrow2+}\frac{(x-2)(x+3)}{x-2}[/tex]
[tex]lim_{x\rightarrow2+}\frac{(2-2)(2+3)}{2-2}[/tex] = undefined (?)
(x approaching 2 from the negative side)
[tex]lim_{x\rightarrow2-}\frac{(-x+2)(x+3)}{x-2}[/tex]
[tex]lim_{x\rightarrow2-}\frac{(-2+2)(2+3)}{2-2}[/tex] = undefined
Why do I always get devision by zero and undefined, did I do somethin wrong?
b) [tex]lim_{x\rightarrow0}\frac{\sqrt{4-x}-2}{\sqrt{x+1}-1}[/tex]
Because we have a sqrt on the denominator we mulitply top & bottom by the conjugate and we rationalize the denominator,
[tex]lim_{x\rightarrow0}\frac{(\sqrt{4-x}-2)(\sqrt{x+1}+1)}{(\sqrt{x+1}-1)\sqrt{x+1}+1)}[/tex]
The denominator expands and simplifies to:
[tex]lim_{x\rightarrow0}\frac{(\sqrt{4-x}-2)(\sqrt{x+1}+1)}{x+1-1}[/tex]
What can i do next? should I expand the numerator too? ( it would become
[tex]\sqrt{4-x.x+1}+\sqrt{4-x}-2\sqrt{x+1}-2[/tex])
But I don't know how to do this one.