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How do you find the limit #(x+5)(1/(2x)+1/(x+2))# as #x->0^+#?

Answer 1

Ezekiel Alexander

#lim_(x->0^+) (x+5)(1/(2x)+1/(x+2)) = +oo#

We can write #f(x)# as:

#(x+5)(1/(2x)+1/(x+2)) = (x+5)/(2x) +(x+5)/(x+2) = 1/2+5/(2x)+(x+5)/(x+2)#

For #x->0^+# obviously the first and last addendum are finite, but #5/(2x) -> +oo#, thus:

#lim_(x->0^+) (x+5)(1/(2x)+1/(x+2)) = +oo#

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Answer 2

Ezekiel Alexander

To find the limit of the expression (x+5)(1/(2x)+1/(x+2)) as x approaches 0 from the positive side, we can simplify the expression first. By multiplying the terms, we get (x+5)(1/(2x)+1/(x+2)) = (x+5)(x+2+2x)/(2x(x+2)). Simplifying further, we have (x+5)(3x+2)/(2x(x+2)).

Now, we can evaluate the limit by substituting 0 for x in the simplified expression. Plugging in x=0, we get (0+5)(3(0)+2)/(2(0)(0+2)). Simplifying this, we have 5(2)/(0), which is undefined. Therefore, the limit of the expression as x approaches 0 from the positive side does not exist.

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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;0^+#?

How do you find the limit #(x+5)(1/(2x)+1/(x+2))# as #x->0^+#?