Evaluate the limit using L'Hospital's rule
lim
x
→
0
e
x
+
3
x
−
1
4
x
\displaystyle \lim_{{{x}\to{0}}}{\frac{{{e}^{{x}}+{3}{x}-{1}}}{{{4}{x}}}}
x
→
0
lim
4
x
e
x
+
3
x
−
1
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\displaystyle
Enter your answer as a number (like 5, -3, 2.2172) or as a calculation (like 5/3, 2^3, 5+4)
Enter DNE for Does Not Exist, oo for Infinity