# How do you find the limit of #(x+5)/(25-x^2)# as x approaches #5^-#?

It is

#lim_(x->5^-) (x+5)/(25-x^2)=lim_(x->5^-) (x+5)/((5-x)*(5+x))= lim_(x->5^-) cancel(x+5)/((5-x)*cancel(5+x))= lim_(x->5^-) 1/(x-5)=-oo#

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To find the limit of (x+5)/(25-x^2) as x approaches 5^-, we can substitute the value of x into the expression. Plugging in x = 5^- gives us (5^- + 5)/(25 - (5^-)^2). Simplifying this expression, we get (5^- + 5)/(25 - 5^-2). As x approaches 5^-, which means x is slightly less than 5, the value of 5^- becomes very small, approaching zero. Therefore, the limit of (x+5)/(25-x^2) as x approaches 5^- is (0 + 5)/(25 - 0^2), which simplifies to 5/25 or 1/5.

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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.

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.

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 of #(arctan(x)) / (5x)# as x approaches 0?
- How do you evaluate the limit #x/(sqrt(3x^2+1)# as x approaches #oo#?
- How do you evaluate the limit #lim (e^t-1)/sint# as #t->0#?
- Evaluate the limit? : #lim_(x rarr 0) ( tanx-x ) / (x-sinx) #
- How do you find the limit of #(2x-1)^3# as #x->0#?

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