Question

determine the limits
a) lim x-> 1 x^2 -9x+14 / x^2 +6x-7
b) lim x->- 2/5 -5x^2-2x/|2+5x|

Answer 1

\[ \large \lim_{x\rightarrow 1}{{x^2-9x+14}\over{x^2+6x-7}}\]

You cannot substitute here because the denominator would be zero when x=1.

Try factoring the numerator and the denominator.

Answer 2

thank you very much sir!!! yes.. i did substituted and got zero. then when i am factorizing i am getting ( x-7)( x+2) / (x+7) (x-1) . so how can conclude here the limit

Answer 3

sorry! (x-7) (x-2) the numerator

Answer 4

You do not get 0/0 but I can't see how to get the x-1 factor out of the denominator.

If you copied the second problem correctly there is no problem. You can substitute x = 2/5 and denominator becomes 4.

Answer 5

can i conclude as limit does not exist for first question?

Answer 6

I think so. No limit for number 1. I will plot it and get back if it seems to have one.

Answer 7

http://fooplot.com/plot/5spmgd4khd

Here's a plot. No limit, and no way to patch it up.

Answer 8

ok thanks sir.. really appreciate the help.

Answer 9

−5−(2/5)2−2(−2/5)|2+5(−2/5)|=−4/5−−4/5 / len(2+5(−2/5))
= 0

Answer 10

\[\large {{-5x^2 -2x} \over{|2+5x|} } = {-5{4 \over 25} - 2{2 \over 5} \over 4} ={-{4 \over 5} - {4 \over 5} \over 4} = -{1 \over 5} - {1 \over 5}=-{2 \over 5}\]

Answer 11

limit is -(2/5). sorry for giving wrong signs.

Answer 12

Oh, in that case may this one does not exist either. Let me think a while.

Answer 13

http://fooplot.com/plot/e23rb4mbgv

Nope. No limit. See plot.

Answer 14

Sir.. you helped me lots for my assignment :)~thankssssssss

Answer 15

Please if you don't mind, can you explain how can i use fooPlot?