Question
Evaluate x→0limxsin5x−sin3x and use the result to find the derivative of f(x)=sin5x−sin3x at x=0.
Solution — Step by Step
limx→0xsin5x−sin3x=limx→0xsin5x−limx→0xsin3x
Both limits exist (we’ll evaluate them), so we can split.
Recall: u→0limusinu=1. So:
limx→0xsin5x=limx→05xsin5x⋅5=1⋅5=5
Similarly, x→0limxsin3x=3.
limx→0xsin5x−sin3x=5−3=2
By definition, f′(0)=x→0limx−0f(x)−f(0). Here f(0)=0, so f′(0)=x→0limxsin5x−sin3x=2.
Cross-check using the standard differentiation rules: f′(x)=5cos5x−3cos3x, so f′(0)=5−3=2. ✓
Why This Works
The standard limit limu→0sinu/u=1 is the foundation of all trigonometric limit problems. The trick is to manufacture this form: if you have sin(kx) in the numerator with x in the denominator, multiply and divide by k to get sin(kx)/(kx) — which has limit 1 — and pick up a factor of k.
The derivative-at-a-point connection is direct: the limit definition of the derivative is exactly what we evaluated. This is why limit calculations and derivative calculations agree.
Memorize three standard limits:
- limx→0sinx/x=1
- limx→0(1−cosx)/x2=1/2
- limx→0(ex−1)/x=1
These three solve 80% of CBSE/JEE Main limit problems.
Alternative Method
Apply L’Hôpital’s rule (if your syllabus allows): the limit is 0/0 form. Differentiate top and bottom:
limx→015cos5x−3cos3x=5−3=2
Same answer in one step. JEE Main accepts L’Hôpital, but CBSE class 11 expects the standard-limit method.
Students try to expand sin5x=sin4xcosx+cos4xsinx — too messy. Using the standard limit sinu/u→1 is cleaner. Always look for that pattern first.
Final answer: limit =2, f′(0)=2.