limh→0 5ex−5ex+h / 3h =
To find the limit of the given expression as h approaches 0, let’s break it down step by step using algebraic manipulation and properties of exponentials
To find the limit of the given expression as h approaches 0, let’s break it down step by step using algebraic manipulation and properties of exponentials.
Step 1: Simplify the numerator.
Since the numerator has two terms with the same base (e), we can combine the terms using the properties of exponentials: ex / ey = ex-y. Applying this property, we can rewrite the numerator as follows:
5ex – 5ex+h = 5ex (1 – e-h)
Step 2: Simplify the denominator.
The denominator is 3h, and it cannot be simplified further.
Step 3: Substitute the simplified numerator and denominator back into the original expression.
The expression now becomes:
limh→0 5ex (1 – e-h) / 3h
Step 4: Apply the limit as h approaches 0.
We can now evaluate the limit. As h approaches 0, the term e-h approaches 1 since e0 = 1. Therefore, we can rewrite the expression with this substitution:
limh→0 5ex (1 – 1) / 3h
Simplifying further, we have:
limh→0 5ex (0) / 3h
The term 0 in the numerator makes the entire numerator equal to 0. Therefore, we have:
limh→0 0 / 3h
Step 5: Evaluate the limit.
To evaluate the limit, we observe that the denominator approaches 0 as h approaches 0. Since any nonzero number divided by 0 is undefined, the limit is undefined.
Hence, the limit of the given expression as h approaches 0 is undefined.
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