Question

Lim tends to sqrt2.( x^2 - 4 )/ x^2 + 3*x^sqrt 2 - 8

Answer 1

factor and cancel

Answer 2

ok

Answer 3

\[\lim_{x \rightarrow \sqrt{2}}(\frac{x^2-4}{x^2}+3x^\sqrt{2}-8) \\ \]
I really see this
and I'm trying to see something else because I don't think it is this

Answer 4

(X^2 + 3*x* sqrt2 -8) is the denominator.apologies for not writing clearly. Thanks in advance for your help!!

Answer 5

\[\lim_{x \rightarrow \sqrt{2}}\frac{x^2-4}{x^2+3 \sqrt{2} x-8}\]
well you will have -2/0 if you plug in sqrt(2)
so your limit could be -infinity or +infinity or does not exist
and technically you could say -inf or +inf doesn't exist
but most like you to be more specif when and if you can

let's consider sqrt(2) from the right

\[\lim_{x \rightarrow \sqrt{2}^+}\frac{x^2-4}{x^2+3 \sqrt{2} x-8}\]
that means we are looking at values slightly greater than sqrt(2)
so we have the following inequality to play with:
\[x> \sqrt{2}\]
so squaring both sides
\[x^2 > 2 \]
subtracting 4 on both sides
\[x^2-4>-2\]
x^2-4 is nearly greater than -2 for values of x near sqrt(2) from the right
basically this means we will consider the factor (x^2-4) negativee for values near x=sqrt(2) from the right

so now lets look at the bottom
we already have
\[x>\sqrt{2} \\ \text{ so we have } 3 \sqrt{2}x>3 \sqrt{2} \sqrt{2} \text{ if we multiply both sides by} 3 \sqrt{2} \\ \text{ which means } 3 \sqrt{2} x>6 \\ \text{ so for the bottom factor we have } \\ x^2+3 \sqrt{2}x-8>2+6-8=0\]
so since the bottom is greater than 0
we will consider the factor on bottom positive for values of x near sqrt(2) from the right

so we have the following:
\[\frac{\text{ negative }}{ \text{ positive }} \infty =-\infty=\lim_{x \rightarrow \sqrt{2}^+}\frac{x^2-4}{x^2+3 \sqrt{2} x-8}\]

Answer 6

consider the left hand limit

Answer 7

you can do it in a similar way

Answer 8

Hi freckles""" many many thanks !!! But the teacher gave the answer as 8/5!!! Pls help

Answer 9

Will if you really meant
\[\lim_{x \rightarrow \sqrt{2}}\frac{x^2-4}{x^2+3 \sqrt{2} x-8}\]
the limit does not exist

Answer 10

http://www.wolframalpha.com/input/?i=limit%28+%28x%5E2-4%29%2F%28x%5E2%2B3+sqrt%282%29+x-8%29%2Cx%3Dsqrt%282%29%29
Other sources can also tell you this.

Answer 11

maybe you meant this instead:
http://www.wolframalpha.com/input/?i=limit%28+%28x%5E2-2%29%2F%28x%5E2%2B3+sqrt%282%29+x-8%29%2Cx%3Dsqrt%282%29%29
but this still doesn't give the result 8/5
it gives 2/5

Answer 12

Hi freckles noted the answer 2/5 but in that case the numerator in the problem is x^2 - 2 whereas in our case the numerator is x^2 -4. Would it mean something and answer is 8/5???