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How do you find the limit #(x+5)(1/(2x)+1/(x+2))# as #x->1#?

Answer 1

#31/6#

When #lim(x →1)#, we essentially just want to be able to plug in 1 in the function. Sometimes this causes a need to rearrange, but here we can just plug it in straight away.

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Answer 2

Charles Anctil

To find the limit of the expression (x+5)(1/(2x)+1/(x+2)) as x approaches 1, we can simplify the expression first. By combining the fractions and simplifying, we get (x+5)((x+2+2x)/(2x(x+2))). Next, we can cancel out common factors and simplify further. After canceling out the common factor of (x+2), we are left with (x+5)(1/(2x)). Finally, we substitute x=1 into the simplified expression to find the limit. Plugging in x=1, we get (1+5)(1/(2(1))) = 6/2 = 3. Therefore, the limit of the expression as x approaches 1 is 3.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you find the limit #(x+5)(1/(2x)+1/(x+2))# as #x-&gt;1#?

How do you find the limit #(x+5)(1/(2x)+1/(x+2))# as #x->1#?