# How do you find the limit of #3/sqrt(x-5) # as x approaches #5^+#?

Given,

By signing up, you agree to our Terms of Service and Privacy Policy

To find the limit of 3/sqrt(x-5) as x approaches 5^+, we can substitute the value of 5 into the expression. However, this would result in division by zero, which is undefined. Therefore, the limit does not exist.

By signing up, you agree to our Terms of Service and Privacy Policy

To find the limit of ( \frac{3}{\sqrt{x-5}} ) as ( x ) approaches ( 5^+ ), we substitute ( x = 5 ) into the expression:

[ \lim_{{x \to 5^+}} \frac{3}{\sqrt{x-5}} ]

[ = \frac{3}{\sqrt{5-5}} ]

[ = \frac{3}{\sqrt{0}} ]

[ = \frac{3}{0^+} ]

Since the denominator approaches zero from the positive side, the overall expression tends towards positive infinity. Therefore, the limit is ( +\infty ).

By signing up, you agree to our Terms of Service and Privacy Policy

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.

- What is the limit as x approaches 0 of #sin(1/x)#?
- How do you use l'Hopital's Rule to evalute #lim_(xrarroo) (x+sin(x))/x#?
- Why is the limit as #x->2# for #3/(2-x)# undefined?
- How do you find the limit of # 1/(x(1 + x²))# as x approaches #0^-#?
- How do you prove the statement lim as x approaches 0 for #x^2 = 0# using the epsilon and delta definition?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7