How would i solve these limits algebraically?

Let M be a maximum of (1/x). Let N = (1/2M). Then

Then (1/N) = 2M

As x->(0+ it reaches N.

For x = (1/N), (1/x) = 2M. Since 2M > M, M cannot be the maximum in

1) lim x->0+ 1/x answer is + ∞

Therefore there is no maximum of (1/x) as x->0+ and the limit of (1/x) as x->0+

is unlimited and therefore must be +1) lim x->0+ 1/x answer is + ∞.

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2) lim x-> 0+ 1/(x^2-x) answer is - ∞

Suppose that (1 / (x^2 - x)) is negative and limited by a finite magnitude.

Let M be such that

limit (1 / (x^2 - x) = M

x->0+

Then consider the specific x such that

limit (1 / x^2 - x) = M

x->0+

Let z = 2x

Then

limit (1 / ((z)^2 - z) =

x->0+

limit (1 / ((2x)^2 - 2x) >

x->0+

limit (1/2) *( (1 / ((x)^2 - x) and is also negative.

Hence the hypothesis of the existence of M is false and the limit is negative and unlimited in magnitude, in other words, -∞

in that case just plug in somesmall number that is close to it and see what happens.

in first part you have x->0+ this means that x>0 but very close to it (so you can try x=0.001 or x=0.000000000001 or x=0.0000000000000000000000001 etc.)

it is worth mentioning that because x>0 result will be positive since 1/(small positive number) is also positive but large number.

also note that x^2 grows fast if x>1 but it becomes smaller if x<1.

for example 2^2=4, 3^2=9 etc. (growing)

but

0.5^2=0.25; 0.25^2=0.0625 (decreasing)

also note that any number squared is positive

this is why in second example you have denominator negative since magnitude of x is greater than x^2 (because x is small)

finally note that if you have x-->0- then you need to use negative x value