Mixing
Find the limit of $lim_{ xto 14} frac{x^2-14x}{x^2-196}$
$begingroup$
Find the following limit when $$lim_{ xto 14} frac{x^2-14x}{x^2-196}$$ I have find that subbing in 14 gives a null answer. But am having trouble factorizing this equation
calculus
edited Apr 10 '15 at 4:49
RE60K
14k22155
asked Apr 10 '15 at 4:46
TinaTina
91
$endgroup$
add a comment |
$begingroup$
Find the following limit when $$lim_{ xto 14} frac{x^2-14x}{x^2-196}$$ I have find that subbing in 14 gives a null answer. But am having trouble factorizing this equation
calculus
edited Apr 10 '15 at 4:49
RE60K
14k22155
asked Apr 10 '15 at 4:46
TinaTina
91
$endgroup$
$begingroup$
Try L'Hôpital's rule
$endgroup$
– Kevin
Apr 10 '15 at 4:49
2
$begingroup$
@Kevin it's too overboard.
$endgroup$
– RE60K
Apr 10 '15 at 4:49
add a comment |
$begingroup$
Find the following limit when $$lim_{ xto 14} frac{x^2-14x}{x^2-196}$$ I have find that subbing in 14 gives a null answer. But am having trouble factorizing this equation
calculus
edited Apr 10 '15 at 4:49
RE60K
14k22155
asked Apr 10 '15 at 4:46
TinaTina
91
$endgroup$
Find the following limit when $$lim_{ xto 14} frac{x^2-14x}{x^2-196}$$ I have find that subbing in 14 gives a null answer. But am having trouble factorizing this equation
calculus
calculus
edited Apr 10 '15 at 4:49
RE60K
14k22155
asked Apr 10 '15 at 4:46
TinaTina
91
edited Apr 10 '15 at 4:49
RE60K
14k22155
asked Apr 10 '15 at 4:46
TinaTina
91
edited Apr 10 '15 at 4:49
RE60K
14k22155
edited Apr 10 '15 at 4:49
RE60K
14k22155
edited Apr 10 '15 at 4:49
RE60K
14k22155
14k22155
asked Apr 10 '15 at 4:46
TinaTina
91
asked Apr 10 '15 at 4:46
TinaTina
91
asked Apr 10 '15 at 4:46
TinaTina
91
91
$begingroup$
Try L'Hôpital's rule
$endgroup$
– Kevin
Apr 10 '15 at 4:49
2
$begingroup$
@Kevin it's too overboard.
$endgroup$
– RE60K
Apr 10 '15 at 4:49
add a comment |
$begingroup$
Try L'Hôpital's rule
$endgroup$
– Kevin
Apr 10 '15 at 4:49
2
$begingroup$
@Kevin it's too overboard.
$endgroup$
– RE60K
Apr 10 '15 at 4:49
$begingroup$
Try L'Hôpital's rule
$endgroup$
– Kevin
Apr 10 '15 at 4:49
$begingroup$
Try L'Hôpital's rule
$endgroup$
– Kevin
Apr 10 '15 at 4:49
2
2
$begingroup$
@Kevin it's too overboard.
$endgroup$
– RE60K
Apr 10 '15 at 4:49
$begingroup$
@Kevin it's too overboard.
$endgroup$
– RE60K
Apr 10 '15 at 4:49
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
$$dfrac{x^2-14x}{x^2-196}=dfrac{x(x-14)}{(x-14)(x+14)}$$
answered Apr 10 '15 at 4:50
BumblebeeBumblebee
9,74612551
$endgroup$
add a comment |
$begingroup$
$$frac{x^2-14x}{x^2-196}=frac{x(x-14)}{(x+14)(x-14)}={xover x+14}$$ Then you will find the limit$={14over28}={1over2}$.
edited Jan 22 at 3:42
Math Lover
15910
answered Apr 10 '15 at 4:51
FreyFrey
648312
$endgroup$
add a comment |
$begingroup$
$$x^2-14x=x(x-14)text{ and } x^2-196=x^2-14^2=(x-14)(x+14)$$
answered Apr 10 '15 at 4:50
RE60KRE60K
14k22155
$endgroup$
add a comment |
$begingroup$
Since x tends to 14 and when you replace x with 14 you get 0/0 condition. Apply L'hospital rule i.e. differentiate both numerator and denominator w.r.t. x. Apply the limit and you have your answer.
answered Apr 10 '15 at 7:44
AjayAjay
897
$endgroup$
add a comment |
$begingroup$
Hint: when $x neq 14$ we have
$$
frac{x^2-14x}{x^2-196} = frac{x(x-14)}{(x+14)(x-14)} = frac{x}{x+14}.
$$
answered Apr 10 '15 at 4:50
Yuval FilmusYuval Filmus
48.7k472145
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$dfrac{x^2-14x}{x^2-196}=dfrac{x(x-14)}{(x-14)(x+14)}$$
answered Apr 10 '15 at 4:50
BumblebeeBumblebee
9,74612551
$endgroup$
add a comment |
$begingroup$
$$dfrac{x^2-14x}{x^2-196}=dfrac{x(x-14)}{(x-14)(x+14)}$$
answered Apr 10 '15 at 4:50
BumblebeeBumblebee
9,74612551
$endgroup$
add a comment |
$begingroup$
$$dfrac{x^2-14x}{x^2-196}=dfrac{x(x-14)}{(x-14)(x+14)}$$
answered Apr 10 '15 at 4:50
BumblebeeBumblebee
9,74612551
$endgroup$
$$dfrac{x^2-14x}{x^2-196}=dfrac{x(x-14)}{(x-14)(x+14)}$$
answered Apr 10 '15 at 4:50
BumblebeeBumblebee
9,74612551
answered Apr 10 '15 at 4:50
BumblebeeBumblebee
9,74612551
answered Apr 10 '15 at 4:50
BumblebeeBumblebee
9,74612551
answered Apr 10 '15 at 4:50
BumblebeeBumblebee
9,74612551
9,74612551
add a comment |
add a comment |
$begingroup$
$$frac{x^2-14x}{x^2-196}=frac{x(x-14)}{(x+14)(x-14)}={xover x+14}$$ Then you will find the limit$={14over28}={1over2}$.
edited Jan 22 at 3:42
Math Lover
15910
answered Apr 10 '15 at 4:51
FreyFrey
648312
$endgroup$
add a comment |
$begingroup$
$$frac{x^2-14x}{x^2-196}=frac{x(x-14)}{(x+14)(x-14)}={xover x+14}$$ Then you will find the limit$={14over28}={1over2}$.
edited Jan 22 at 3:42
Math Lover
15910
answered Apr 10 '15 at 4:51
FreyFrey
648312
$endgroup$
add a comment |
$begingroup$
$$frac{x^2-14x}{x^2-196}=frac{x(x-14)}{(x+14)(x-14)}={xover x+14}$$ Then you will find the limit$={14over28}={1over2}$.
edited Jan 22 at 3:42
Math Lover
15910
answered Apr 10 '15 at 4:51
FreyFrey
648312
$endgroup$
$$frac{x^2-14x}{x^2-196}=frac{x(x-14)}{(x+14)(x-14)}={xover x+14}$$ Then you will find the limit$={14over28}={1over2}$.
edited Jan 22 at 3:42
Math Lover
15910
answered Apr 10 '15 at 4:51
FreyFrey
648312
edited Jan 22 at 3:42
Math Lover
15910
edited Jan 22 at 3:42
Math Lover
15910
edited Jan 22 at 3:42
Math Lover
15910
15910
answered Apr 10 '15 at 4:51
FreyFrey
648312
answered Apr 10 '15 at 4:51
FreyFrey
648312
answered Apr 10 '15 at 4:51
FreyFrey
648312
648312
add a comment |
add a comment |
$begingroup$
$$x^2-14x=x(x-14)text{ and } x^2-196=x^2-14^2=(x-14)(x+14)$$
answered Apr 10 '15 at 4:50
RE60KRE60K
14k22155
$endgroup$
add a comment |
$begingroup$
$$x^2-14x=x(x-14)text{ and } x^2-196=x^2-14^2=(x-14)(x+14)$$
answered Apr 10 '15 at 4:50
RE60KRE60K
14k22155
$endgroup$
add a comment |
$begingroup$
$$x^2-14x=x(x-14)text{ and } x^2-196=x^2-14^2=(x-14)(x+14)$$
answered Apr 10 '15 at 4:50
RE60KRE60K
14k22155
$endgroup$
$$x^2-14x=x(x-14)text{ and } x^2-196=x^2-14^2=(x-14)(x+14)$$
answered Apr 10 '15 at 4:50
RE60KRE60K
14k22155
answered Apr 10 '15 at 4:50
RE60KRE60K
14k22155
answered Apr 10 '15 at 4:50
RE60KRE60K
14k22155
answered Apr 10 '15 at 4:50
RE60KRE60K
14k22155
14k22155
add a comment |
add a comment |
$begingroup$
Since x tends to 14 and when you replace x with 14 you get 0/0 condition. Apply L'hospital rule i.e. differentiate both numerator and denominator w.r.t. x. Apply the limit and you have your answer.
answered Apr 10 '15 at 7:44
AjayAjay
897
$endgroup$
add a comment |
$begingroup$
Since x tends to 14 and when you replace x with 14 you get 0/0 condition. Apply L'hospital rule i.e. differentiate both numerator and denominator w.r.t. x. Apply the limit and you have your answer.
answered Apr 10 '15 at 7:44
AjayAjay
897
$endgroup$
add a comment |
$begingroup$
Since x tends to 14 and when you replace x with 14 you get 0/0 condition. Apply L'hospital rule i.e. differentiate both numerator and denominator w.r.t. x. Apply the limit and you have your answer.
answered Apr 10 '15 at 7:44
AjayAjay
897
$endgroup$
Since x tends to 14 and when you replace x with 14 you get 0/0 condition. Apply L'hospital rule i.e. differentiate both numerator and denominator w.r.t. x. Apply the limit and you have your answer.
answered Apr 10 '15 at 7:44
AjayAjay
897
answered Apr 10 '15 at 7:44
AjayAjay
897
answered Apr 10 '15 at 7:44
AjayAjay
897
answered Apr 10 '15 at 7:44
AjayAjay
897
897
add a comment |
add a comment |
$begingroup$
Hint: when $x neq 14$ we have
$$
frac{x^2-14x}{x^2-196} = frac{x(x-14)}{(x+14)(x-14)} = frac{x}{x+14}.
$$
answered Apr 10 '15 at 4:50
Yuval FilmusYuval Filmus
48.7k472145
$endgroup$
add a comment |
$begingroup$
Hint: when $x neq 14$ we have
$$
frac{x^2-14x}{x^2-196} = frac{x(x-14)}{(x+14)(x-14)} = frac{x}{x+14}.
$$
answered Apr 10 '15 at 4:50
Yuval FilmusYuval Filmus
48.7k472145
$endgroup$
add a comment |
$begingroup$
Hint: when $x neq 14$ we have
$$
frac{x^2-14x}{x^2-196} = frac{x(x-14)}{(x+14)(x-14)} = frac{x}{x+14}.
$$
answered Apr 10 '15 at 4:50
Yuval FilmusYuval Filmus
48.7k472145
$endgroup$
Hint: when $x neq 14$ we have
$$
frac{x^2-14x}{x^2-196} = frac{x(x-14)}{(x+14)(x-14)} = frac{x}{x+14}.
$$
answered Apr 10 '15 at 4:50
Yuval FilmusYuval Filmus
48.7k472145
answered Apr 10 '15 at 4:50
Yuval FilmusYuval Filmus
48.7k472145
answered Apr 10 '15 at 4:50
Yuval FilmusYuval Filmus
48.7k472145
answered Apr 10 '15 at 4:50
Yuval FilmusYuval Filmus
48.7k472145
48.7k472145
add a comment |
add a comment |
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$begingroup$
Try L'Hôpital's rule
$endgroup$
– Kevin
Apr 10 '15 at 4:49
2
$begingroup$
@Kevin it's too overboard.
$endgroup$
– RE60K
Apr 10 '15 at 4:49