# How do you find the limit of #(5x)/sqrt(x+2)# as #x->7#?

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

To find the limit of (5x)/sqrt(x+2) as x approaches 7, we can use direct substitution. Plugging in x=7 into the expression, we get (5*7)/sqrt(7+2), which simplifies to 35/sqrt(9). The square root of 9 is 3, so the expression becomes 35/3. Therefore, the limit of (5x)/sqrt(x+2) as x approaches 7 is 35/3.

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.

- For what values of x, if any, does #f(x) = 1/((x-4)(x-8)) # have vertical asymptotes?
- What is #lim_(x->0) (1-cosx)^2/x # ?
- How do you find the limit of #(-x+4)# as x approaches -3?
- How do you find #lim (1-2t^-1+t^-2)/(3-4t^-1)# as #t->0#?
- How do you prove that the limit of #5 - 2x# as x approaches 2 is equal to 1 using the epsilon delta proof?

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